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Sequences and Series

3.2 Series

33 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

Write out the first five partial sums corresponding to the series ∑n=1∞1n\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}.

You don't need to simplify the terms.

Hint

SNS_N is the sum of the terms corresponding to n=1n=1 through n=Nn=N.

Answer

NSN1121+1231+12+1341+12+13+1451+12+13+14+15\begin{array}{l|l} \mathbf{N}&\mathbf{S_N}\\ \hline 1 & 1\\[7pt] 2 & 1+\frac12\\[7pt] 3 & 1+\frac12 + \frac13\\[7pt] 4 & 1+\frac12 + \frac13+\frac14\\[7pt] 5 & 1+\frac12 + \frac13+\frac14+\frac15\\[7pt] \end{array}

Full solution

The NNth term of the sequence of partial sums, SNS_N, is the sum of the first NN terms of the series ∑n=1∞1n\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}.

NSN1121+1231+12+1341+12+13+1451+12+13+14+15\begin{array}{l|l} \mathbf{N}&\mathbf{S_N}\\ \hline 1 & 1\\[7pt] 2 & 1+\frac12\\[7pt] 3 & 1+\frac12 + \frac13\\[7pt] 4 & 1+\frac12 + \frac13+\frac14\\[7pt] 5 & 1+\frac12 + \frac13+\frac14+\frac15\\[7pt] \end{array}
Q2Stage 1

Every student who comes to class brings their instructor cookies, and leaves them on the instructor's desk. Let CkC_k be the total number of cookies on the instructor's desk after the kkth student comes.

If C11=20C_{11}=20, and C10=17C_{10}=17, how many cookies did the 11th student bring to class?

Hint

Note CkC_k is the cumulative number of cookies.

Answer

3

Full solution

If there were a total of 17 cookies before Student 11 came, and 20 cookies after, then Student 11 brought 3 cookies.

Figure from prob_s3.2, line 2

Figure from prob_s3.2, line 2

Q3Stage 1

Suppose the sequence of partial sums of the series ∑n=1∞an\displaystyle\sum_{n=1}^\infty a_n is {SN}={NN+1}\{S_N\} = \left\{\dfrac{N}{N+1}\right\}.

  1. What is {an}\{a_n\}?

  2. What is lim⁡n→∞an\lim\limits_{n \to \infty} a_n?

  3. Evaluate ∑n=1∞an\displaystyle\sum_{n=1}^\infty a_n.

Hint

How is (a) related to Question 2?

Answer

(a) an={12 if n=11n(n+1) else \displaystyle a_n = \begin{cases} \frac12 &\text{ if } n=1\\ \dfrac{1}{n(n+1)} &\text{ else }\\ \end{cases} (b) 0 (c) 1

Full solution

  1. We find {an}\{a_n\} from {SN}\{S_N\} using the same logic as Question 2. SNS_N is the sum of the first NN terms of {an}\{a_n\}, and SN−1S_{N-1} is the sum of all the same terms except aNa_N. So, aN=SN−SN−1a_N = S_N-S_{N-1} when N≥2N \geq 2. Written another way:

    SN=a1+a2+a3+⋯+aN−2+aN−1+aNSN−1=a1+a2+a3+⋯+aN−2+aN−1\begin{align*}\color{blue}S_N&\color{blue}=a_1+a_2+a_3+\cdots+a_{N-2}+a_{N-1}+a_N\\ \color{red}S_{N-1}&\color{red}=a_1+a_2+a_3+\cdots+a_{N-2}+a_{N-1}\end{align*}

    So,

    SN−SN−1=[a1+a2+a3+⋯+aN−2+aN−1+aN]−[a1+a2+a3+⋯+aN−2+aN−1]=aN\begin{align*}\color{blue}S_N -\color{red}S_{N-1} &=\color{blue}\Big[a_1+a_2+a_3+\cdots+a_{N-2}+a_{N-1}+a_N\Big]\\&- \color{red}\Big[a_1+a_2+a_3+\cdots+a_{N-2}+a_{N-1} \Big]\\ &=\color{blue}a_N\end{align*}

    So, we calculate

    aN=SN−SN−1=(NN+1)−(N−1N−1+1)=N2N(N+1)−N2−1N(N+1)=1N(N+1)\begin{align*}a_N&=S_N-S_{N-1}=\left(\dfrac{N}{N+1}\right)-\left(\dfrac{N-1}{N-1+1}\right)\\ &=\frac{N^2}{N(N+1)}-\frac{N^2-1}{N(N+1)}\\ &=\frac{1}{N(N+1)}\end{align*}

    Therefore,

    an=1n(n+1)\begin{align*}a_n&=\frac{1}{n(n+1)}\end{align*}

    Remark: the formula given for SNS_N has S0=0S_0=0, which makes sense: the sum of no terms at all should be 0. However, it is common for a sequence of partial sums to start at N=1N=1. (This fits our definition of a partial sum–we don't really define the “sum of no terms.") In this case, a1a_1 must be calculated separately from the other terms of {an}\{a_n\}. To find a1a_1, we simply set a1=S1a_1=S_1, which (to reiterate) might not be the same as S1−S0S_1-S_0.

  2. lim⁡n→∞an=lim⁡n→∞1n(n+1)=0.\lim\limits_{n \to \infty} a_n =\lim\limits_{n \to \infty} \dfrac{1}{n(n+1)}=0.

    That is, the terms we're adding up are getting very, very small as we go along.

  3. By Definition 3.2.3 in the CLP-2 text,

    ∑n=1∞an=lim⁡N→∞SN=lim⁡N→∞NN+1=1\sum_{n=1}^\infty a_n = \lim_{N \to \infty}S_N = \lim_{N \to \infty}\dfrac{N}{N+1}=1

    That is, as we add more and more terms of our series, our cumulative sum gets very, very close to 1.

Q4Stage 1

Suppose the sequence of partial sums of the series ∑n=1∞an\displaystyle\sum_{n=1}^\infty a_n is {SN}={(−1)N+1N}\{S_N\} = \left\{(-1)^N+\dfrac{1}{N}\right\}.

What is {an}\{a_n\}?

Hint

You'll have to calculate a1a_1 separately from the other terms.

Answer

an={0 if n=12(−1)n−1n(n−1) elsea_n=\begin{cases} 0 & \text{ if } n=1\\ 2(-1)^{n} -\frac{1}{n(n-1)} &\text{ else} \end{cases}

Full solution

As in Question 3,

aN=SN−SN−1=[(−1)N+1N]−[(−1)N−1+1N−1]=(−1)N−(−1)N−1+1N−1N−1=(−1)N+(−1)N+N−1N(N−1)−NN(N−1)=2(−1)N−1N(N−1)\begin{align*}a_N&=S_N-S_{N-1}=\left[(-1)^N+\dfrac{1}{N}\right]-\left[(-1)^{N-1}+\dfrac{1}{N-1}\right]\\ &=(-1)^N-(-1)^{N-1} + \frac{1}{N}-\frac{1}{N-1}\\ &=(-1)^N+(-1)^{N} + \frac{N-1}{N(N-1)}-\frac{N}{N(N-1)}\\ &=2(-1)^{N} -\frac{1}{N(N-1)}\end{align*}

Note, however, that aNa_N is only the same as SN−SN−1S_N-S_{N-1} when N≥2N \geq 2: otherwise, we're trying to calculate S1−S0S_1-S_0, but S0S_0 is not defined. So, we find a1a_1 separately:

a1=S1=(−1)1+11=0\begin{align*}a_1&=S_1=(-1)^1+\frac{1}{1}=0\end{align*}

All together:

an={0 if n=12(−1)n−1n(n−1) else\begin{align*}a_n&=\begin{cases} 0 & \text{ if } n=1\\ 2(-1)^{n} -\frac{1}{n(n-1)} &\text{ else} \end{cases}\end{align*}
Q5Stage 1

Let f(N)f(N) be a formula for the NNth partial sum of ∑n=1∞an\displaystyle\sum_{n=1}^\infty a_n. (That is, f(N)=SNf(N)=S_N.) If f′(N)<0f'(N)<0 for all N>1N > 1, what does that say about ana_n?

Hint

When does adding a number decrease the total sum?

Answer

an<0a_n < 0 for all n≥2n \geq 2

Full solution

If f′(N)<0f'(N)<0, that means f(N)f(N) is decreasing. So, adding more terms makes for a smaller sum. That means the terms we're adding are negative. That is, an<0a_n < 0 for all n≥2n \geq 2.

Questions 6 through 8 invite you to explore geometric sums in a geometric way. This is complementary to than the algebraic method discussed in the text.

Q6Stage 1

Suppose the triangle outlined in red in the picture below has area one.

Figure from prob_s3.2, line 1

Figure from prob_s3.2, line 1

  1. Express the combined area of the black triangles as a series, assuming the pattern continues forever.

  2. Evaluate the series using the picture (not the formula from your book).

Hint

For (b), imagine cutting up the triangle into its black and white parts, then sharing it equally among a certain number of friends. What is the easiest number of friends to share with, making sure each has the same area in their pile?

Answer

(a) ∑n=1∞24n\displaystyle\sum_{n=1}^\infty \frac{2}{4^n} (b) 23\dfrac23

Full solution

(a) To generate the pattern, we repeat the following steps:

  • divide the top triangle into four triangles of equal area,

  • colour the bottom two of them black, and

  • leave the middle one white.

Every time we repeat this sequence, we divide up a triangle with an area one-quarter the size of our previous triangle, and take two of the four resulting pieces. So, our area should end up as a geometric sum with common ratio r=14r=\dfrac14, and coefficient a=2a=2. This is shown more explicitly below.

Since the entire triangle (outlined in red) has area 1, the four smaller triangles below each have area 14\dfrac14. The two black triangles will be added to our total black area; the blue triangle will be subdivided.

Figure from prob_s3.2, line 1

Figure from prob_s3.2, line 1

The blue triangle had area 14\dfrac{1}{4}, so each of the small black triangles below has area (14)(14)=142\left(\dfrac14\right)\left(\dfrac14\right)=\dfrac{1}{4^2}.

Figure from prob_s3.2, line 1

Figure from prob_s3.2, line 1

Each time we make another subdivision, we add two black triangles, each with 14\dfrac{1}{4} the area of the previous black triangles. So, our total black area is:

2(14)+2(142)+2(143)+2(144)+⋯=∑n=1∞24n2\left(\frac14\right)+2\left(\frac{1}{4^2}\right)+2\left(\frac{1}{4^3}\right)+2\left(\frac{1}{4^4}\right)+\cdots = \sum_{n=1}^\infty \frac{2}{4^n}

(b) To evalutate the series, we imagine gathering up all our little triangles and sorting them into three identical piles: the bottom three triangles go in three different piles, the three triangles directly above them go in three different piles, etc. (In the picture below, different colours correspond to different piles.)

Figure from prob_s3.2, line 1

Figure from prob_s3.2, line 1

Since the piles all have equal area, each pile has a total area of 13\dfrac 13. The black area shaded in the problem corresponds to two piles (red and blue above), so

∑n=1∞24n=23\sum_{n=1}^\infty \frac{2}{4^n}=\frac23
Q7Stage 1

Suppose the square outlined in red in the picture below has area one.

Figure from prob_s3.2, line 1

Figure from prob_s3.2, line 1

  1. Express the combined area of the black squares as a series, assuming the pattern continues forever.

  2. Evaluate the series using the picture (not the formula from your book).

Hint

Compare to Question 6.

Answer

(a) ∑n=1∞19n\displaystyle\sum_{n=1}^\infty \frac{1}{9^n} (b) 18\dfrac{1}{8}

Full solution

(a) The pattern can be described as follows: divide the innermost square into 9 equal parts (a 3×33\times 3 grid), choose one square to be black, and another square to subdivide.

The area of the red (outermost) square is 1, so the area of the largest black square is 19\dfrac19. The area of the central, blue square below is also 19\dfrac19.

Figure from prob_s3.2, line 1

Figure from prob_s3.2, line 1

When we subdivide the blue square, the subdivisions each have one-ninth its area, or 192\dfrac{1}{9^2}.

Figure from prob_s3.2, line 1

Figure from prob_s3.2, line 1

We continue taking squares that are one-ninth the area of the previous square. So, our total black area is

19+192+193+⋯=∑n=1∞19n\frac{1}{9}+\frac{1}{9^2}+\frac{1}{9^3}+\cdots = \sum_{n=1}^\infty \frac{1}{9^n}

(b) If we cut up this square along the marks, we can easily share it equally among 8 friends: there are eight squares of area 19\dfrac19 along the outer ring, eight squares of area 192\dfrac1{9^2} along the next ring in, and so on.

Figure from prob_s3.2, line 1

Figure from prob_s3.2, line 1

Since the eight friends all get the same total area, the area each friend gets is 18\dfrac{1}{8}. The area shaded in black in the question corresponds to the pile given to one friend. So,

∑n=1∞19n=18\sum_{n=1}^\infty \frac{1}{9^n}=\frac18
Q8Stage 1

In the style of Questions 6 and 7, draw a picture that represents ∑n=1∞13n\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^n} as an area.

Hint

Iteratively divide a shape into thirds.

Answer

Two possible pictures:

Figure from prob_s3.2, line 2

Figure from prob_s3.2, line 2

Figure from prob_s3.2, line 16

Figure from prob_s3.2, line 16

Full solution

If we start with a shape of area 1, and iteratively divide it into thirds, taking one of the three newly created pieces each time, then the area we take will be equal to the desired series, ∑n=1∞13n\displaystyle\sum_{n=1}^\infty \frac{1}{3^n}.

One way to do this is to start with a rectangle, make three vertical strips, then keep the left strip and subdivide the middle strip.

Figure from prob_s3.2, line 2

Figure from prob_s3.2, line 2

Figure from prob_s3.2, line 30

Figure from prob_s3.2, line 30

We see that the total area we take approaches one-half the total area of the figure, so ∑n=1∞13n=12\displaystyle\sum_{n=1}^\infty \frac{1}{3^n} = \frac12.

Alternately, instead of always taking vertical strips, we could alternate vertical and horizontal slices.

Figure from prob_s3.2, line 3

Figure from prob_s3.2, line 3

Figure from prob_s3.2, line 37

Figure from prob_s3.2, line 37

In this setup, we notice that our strips come in pairs: two large vertical strips, two smaller horizontal strips, two smaller vertical strips, etc. We shaed exactly one of each, so the shaded area is one-half the total area: ∑n=1∞13n=12\displaystyle\sum_{n=1}^\infty \frac{1}{3^n} = \frac12.

Other solutions are possible, as well.

Q9Stage 1

Evaluate ∑n=010015n\displaystyle\sum_{n=0}^{100}\frac{1}{5^n}.

Hint

Equation 3.2.1 in the CLP-2 text tells us ∑n=0Narn=a1−rN+11−r\displaystyle\sum_{n=0}^N ar^n = a\dfrac{1-r^{N+1}}{1-r}, for r≠1r \neq 1.

Answer

5101−14⋅5100\dfrac{5^{101}-1}{4\cdot 5^{100}}

Full solution

Equation 3.2.1 in the CLP-2 text tells us ∑n=0Narn=a1−rN+11−r\displaystyle\sum_{n=0}^N ar^n = a\dfrac{1-r^{N+1}}{1-r}, for r≠1r \neq 1. Our geometric sum has a=1a=1, r=15r=\frac{1}{5}, and N=100N=100. So:

∑n=010015n=1−151011−15=5101−14⋅5100\displaystyle\sum_{n=0}^{100}\frac{1}{5^n} = \frac{1-\frac{1}{5^{101}}}{1-\frac{1}{5}} = \dfrac{5^{101}-1}{4\cdot 5^{100}}
Q10Stage 1

Every student who comes to class brings their instructor cookies, and leaves them on the instructor's desk. Let CkC_k be the total number of cookies on the instructor's desk after the kkth student comes.

If C20=53C_{20}=53, and C10=17C_{10}=17, what does C20−C10=36C_{20}-C_{10}=36 represent?

Hint

Note CkC_k is the cumulative number of cookies.

Answer

All together, there were 36 cookies brought by Student 11 through Student 20.

Full solution

After twenty students have brought their cookies, the pile numbers 53 cookies. 17 of these cookies were brought by students one through ten. So, the remainder (53−17=3653-17=36) is the number of cookies brought by students 11, 12, 13, 14, 15, 16, 17, 18, 19, and 20, together.

Figure from prob_s3.2, line 2

Figure from prob_s3.2, line 2

Q11Stage 1

Evaluate ∑n=5010015n\displaystyle\sum_{n=50}^{100}\frac{1}{5^n}. (Note the starting index.)

Hint

To adjust the starting index, either factor out the first term in the series, or subtract two series. For the subtraction option, consider Question 10.

Answer

551−14⋅5100\dfrac{5^{51}-1}{4\cdot 5^{100}}

Full solution

  • Using the ideas of Question 10, we see:

    ∑n=5010015n=∑n=010015n−∑n=04915n\begin{align*}\sum_{n={50}}^{100} \frac{1}{5^n}&=\sum_{n=0}^{100}\frac{1}{5^n}-\sum_{n=0}^{49}\frac{1}{5^n}\end{align*}

    That is, we want start with the sum of all the terms up to 15100\dfrac{1}{5^{100}}, and then subtract off the ones we actually don't want, which is everything up to 1549\dfrac{1}{5^{49}}. Now, both series are in a form appropriate for Equation 3.2.1 in the CLP-2 text.

    ∑n=010015n−∑n=04915n=1−151011−15−1−15501−15=5101−14⋅5100−550−14⋅549(551551)=5101−14⋅5100−5101−5514⋅5100=551−14⋅5100\begin{align*}\sum_{n=0}^{100}\frac{1}{5^n}-\sum_{n=0}^{49}\frac{1}{5^n}&=\frac{1-\frac{1}{5^{101}}}{1-\frac15}-\frac{1-\frac{1}{5^{50}}}{1-\frac15}\\ &=\frac{5^{101}-1}{4\cdot 5^{100}}-\frac{5^{50}-1}{4\cdot 5^{49}}\left(\frac{5^{51}}{5^{51}}\right)\\ &=\frac{5^{101}-1}{4\cdot 5^{100}}-\frac{5^{101}-5^{51}}{4\cdot 5^{100}}\\ &=\frac{5^{51}-1}{4\cdot 5^{100}}\end{align*}

  • If we write out the first few terms of our series, we see we can factor out a constant to change the starting index.

    ∑n=5010015n=1550+1551+1552+1553+⋯+15100=1550(150+151+152+153+⋯+1550)=∑n=0501550⋅15n\begin{align*}\sum_{n={50}}^{100} \frac{1}{5^n}&= \frac{1}{5^{50}}+ \frac{1}{5^{51}}+\frac{1}{5^{52}}+\frac{1}{5^{53}}+\cdots +\frac{1}{5^{100}}\\ &=\frac{1}{5^{50}}\left(\frac{1}{5^0}+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+\cdots +\frac{1}{5^{50}}\right)\\ &=\sum_{n=0}^{50}\frac{1}{5^{50}}\cdot\frac{1}{5^n}\end{align*}

    Now, our sum is in the form of Equation 3.2.1 in the CLP-2 text with a=1550,a=\dfrac{1}{5^{50}}, r=15r=\dfrac15, and N=50N=50.

    ∑n=0501550⋅15n=1550⋅1−15511−15=1−15514⋅549(551551)=551−14⋅5100\begin{align*}\sum_{n=0}^{50}\frac{1}{5^{50}}\cdot\frac{1}{5^n}&=\frac{1}{5^{50}}\cdot\frac{1-\frac{1}{5^{51}}}{1-\frac15}=\frac{1-\frac{1}{5^{51}}}{4\cdot 5^{49}}\left(\frac{5^{51}}{5^{51}}\right)\\ &=\frac{5^{51}-1}{4\cdot 5^{100}}\end{align*}
Q12Stage 1

  1. Starting on day d=1d=1, every day you give your friend $1d+1\frac{1}{d+1}, and they give $1d\frac{1}{d} back to you. After a long time, how much money have you gained by this arrangement?

  2. Evaluate ∑d=1∞(1d−1(d+1))\displaystyle\sum_{d=1}^\infty \left(\frac{1}{d}-\frac{1}{(d+1)}\right).

  3. Starting on day d=1d=1, every day your friend gives you $(d+1)(d+1), and they take $(d+2)(d+2) from you. After a long time, how much money have you gained by this arrangement?

  4. Evaluate ∑d=1∞((d+1)−(d+2))\displaystyle\sum_{d=1}^\infty \left((d+1)-(d+2)\right).

Hint

Express your gains in (a) and (c) as series.

Answer

(a) As time passes, your gains increase, approaching $1. (b) 1
(c) As time passes, you lose more and more money, without bound. (d) −∞-\infty

Full solution

(a) The table below is a record of our account, with black entries representing the money your friend gives you, and red entries representing the money you give them (which is why the red entries are negative).

d−1d+11dtotal after day d1−121122−1312233−1413344−1514455−1615566−171667\begin{array}{c||c|c||c} \mathbf{d}&\color{red}\mathbf{-\frac{1}{d+1}}&\mathbf{\frac1d}&{\textbf{total after day }d}\\[7pt] \hline 1 & \color{red} -\frac{1}{2} & 1 & \frac{1}{2}\\[7pt] 2&\color{red}-\frac{1}{3} & \frac{1}{2}&\frac{2}{3}\\[7pt] 3&\color{red}-\frac{1}{4} & \frac{1}{3}&\frac{3}{4}\\[7pt] 4&\color{red}-\frac{1}{5} & \frac{1}{4}&\frac{4}{5}\\[7pt] 5&\color{red}-\frac{1}{6} & \frac{1}{5}&\frac{5}{6}\\[7pt] 6&\color{red}-\frac{1}{7} & \frac{1}{6}&\frac{6}{7} \end{array}

After the exchange of day nn, the amount you're left with is $(1−1n+1)\char36 \left(1-\frac{1}{n+1}\right). We see this by the cancellation in the table: the $12\frac12 you gave your friend on day 1 was returned on day 2; the $13\frac13 you gave your friend on day 2 was returned on day 3, etc.

So, after a long time, you'll have gained close to (but always slightly less than) one dollar.

(b) The series ∑d=1∞(1d−1(d+1))\displaystyle\sum_{d=1}^\infty \left(\frac{1}{d}-\frac{1}{(d+1)}\right) describes the scenario in (a), so by our reasoning there,

∑d=1∞(1d−1(d+1))=lim⁡n→∞(1−1n+1)=1\displaystyle\sum_{d=1}^\infty \left(\frac{1}{d}-\frac{1}{(d+1)}\right) = \lim_{n \to \infty}\left(1-\frac{1}{n+1}\right)=1

(c) Again, let's set up an account book.

dd+1−(d+2)total12−3−123−4−234−5−345−6−456−7−567−8−6\begin{array}{c||c|c||c} \mathbf{d}&\mathbf{d+1}&\color{red}\mathbf{-(d+2)}&\textbf{total}\\[7pt] \hline 1 & 2& \color{red} -3 & -1\\[7pt] 2 & 3& \color{red} -4 & -2\\[7pt] 3 & 4& \color{red} -5 & -3\\[7pt] 4 & 5& \color{red} -6 & -4\\[7pt] 5 & 6& \color{red} -7 & -5\\[7pt] 6 & 7& \color{red} -8 & -6\\[7pt] \end{array}

By day dd, you've lost $ dd to your so-called friend. As time goes on, you lose more and more.

(d) The series ∑d=1∞((d+1)−(d+2))\displaystyle\sum_{d=1}^\infty ((d+1)-(d+2)) exactly describes the scenario in part (c), so it diverges to −∞-\infty. You can also see this by writing ∑d=1∞((d+1)−(d+2))=∑d=1∞(−1)=−1−1−1−1−1−⋯\displaystyle\sum_{d=1}^\infty ((d+1)-(d+2)) =\sum_{d=1}^\infty (-1)=-1-1-1-1-1-\cdots.

Be careful to avoid a common mistake with telescoping series: if we look back at our account book, we see that every negative term will cancel with a positive term, with the initial +2+2 as the only term that never cancels. Your friend takes $3, which they return the next day; then they take $4, which they return the next day; then they take $5, which they return the next day, and so on. It's extremely tempting to say that the series adds up to $2, since every other term cancels out eventually. This is where we lean on Definition 3.2.3 in the CLP-2 text: we evaluate the partial sums, which always leave your friend's last withdrawal unreturned. This definition makes sense: saying “I gained two bucks from this exchange" doesn't really capture the reality of your increasing debt.

Q13Stage 1

Suppose ∑n=1∞an=A\displaystyle\sum_{n=1}^\infty a_n = A, ∑n=1∞bn=B\displaystyle\sum_{n=1}^\infty b_n = B, and ∑n=1∞cn=C\displaystyle\sum_{n=1}^\infty c_n = C.

Evaluate ∑n=1∞(an+bn+cn+1)\displaystyle\sum_{n=1}^\infty\left( a_n+b_n+c_{n+1}\right).

Hint

To find the difference between ∑n=1∞cn\displaystyle\sum_{n=1}^\infty c_n and ∑n=1∞cn+1\displaystyle\sum_{n=1}^\infty c_{n+1}, try writing out the first few terms.

Answer

A+B+C−c1A+B+C-c_1

Full solution

Using arithmetic of series, Theorem 3.2.8 in the CLP-2 text, we see

∑n=1∞(an+bn+cn+1)=A+B+∑n=1∞cn+1\sum_{n=1}^\infty\left( a_n+b_n+c_{n+1}\right) = A+B+\sum_{n=1}^\infty c_{n+1}

The question remaining is what do to with the last series. If we write out the terms, we see the difference between ∑n=1∞cn\displaystyle\sum_{n=1}^\infty c_n and ∑n=1∞cn+1\displaystyle\sum_{n=1}^\infty c_{n+1} is simply that the latter is missing c1c_1:

∑n=1∞cn+1=c2+c3+c4+c5+⋯=−c1+c1+c2+c3+c4+c5+⋯=−c1+∑n=1∞cn\begin{align*}\sum_{n=1}^\infty c_{n+1}&=c_2+c_3+c_4+c_5+\cdots\\ &=\textcolor{red}{-c_1+c_1}+c_2+c_3+c_4+c_5+\cdots\\ &=\textcolor{red}{-c_1}+\sum_{n=1}^\infty c_{n}\end{align*}

So,

∑n=1∞(an+bn+cn+1)=A+B+C−c1\begin{align*}\sum_{n=1}^\infty\left( a_n+b_n+c_{n+1}\right)&=A+B+C-\textcolor{red}{c_1}\end{align*}
Q14Stage 1

Suppose ∑n=1∞an=A\displaystyle\sum_{n=1}^\infty a_n = A, ∑n=1∞bn=B≠0\displaystyle\sum_{n=1}^\infty b_n = B \neq 0, and ∑n=1∞cn=C\displaystyle\sum_{n=1}^\infty c_n = C.

True or false: ∑n=1∞(anbn+cn)=AB+C\displaystyle\sum_{n=1}^\infty\left(\dfrac{ a_n}{b_n}+c_{n}\right) = \frac{A}{B}+C.

Hint

You might want to first consider a simpler true or false: ∑n=1∞anbn=?AB\displaystyle\sum_{n=1}^\infty \dfrac{ a_n}{b_n} \stackrel{?}{=} \dfrac{A}{B}.

Answer

in general, false

Full solution

Theorem 3.2.8 in the CLP-2 text, arithmetic of series, doesn't mention division, because in general it doesn't work the way the question suggests. For example, let {an}={bn}=12n\{a_n\}=\{b_n\} = \frac{1}{2^n}. Then:

  • ∑n=0∞an=∑n=0∞bn=11−12=2\sum_{n=0}^{\infty} a_n = \sum_{n=0}^\infty b_n = \frac{1}{1-\frac12}=2, while

  • ∑n=0∞anbn=∑n=0∞1=∞\sum_{n=0}^{\infty}\dfrac{a_n}{b_n} = \sum_{n=0}^{\infty}1 =\infty.

For the statement in the question, we can take {an}={bn}=12n\{a_n\}=\{b_n\} = \frac{1}{2^n}, A=B=2A=B=2, {cn}={0,0,0,…}\{c_n\}=\{0,0,0,\ldots\}, and C=0C=0. We see the statement is false in this case.

So, in general, the statement given is false.

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q15Stage 2Past exam · 2016Q5

To what value does the series 1+13+19+127+181+1243+⋯\displaystyle 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}+ \frac{1}{81}+ \frac{1}{243} + \cdots converge?

Hint

What kind of a series is this?

Answer

32\dfrac{3}{2}

Full solution

We recognize that this is a geometric series:

1+13+19+127+181+1243+⋯=130+131+132+133+134+135+⋅=∑n=0∞13n\begin{align*}1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}+ \frac{1}{81}+ \frac{1}{243} + \cdots&=\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\cdot\\ &=\sum_{n=0}^\infty \frac{1}{3^n}\end{align*}

Using Equation 3.2.2 in the CLP-2 text with r=13r=\dfrac13 and a=1a=1,

=11−13=32.\begin{align*}&=\frac{1}{1-\frac13}=\frac{3}{2}.\end{align*}
Q16Stage 2Past exam · M105 2015A

Evaluate ∑k=7∞18k\displaystyle \sum_{k=7}^{\infty} \frac{1}{8^k}

Hint

This is a special kind of series, that you should recognize.

Answer

17×86\dfrac{1}{7\times 8^6}

Full solution

This is a geometric series, with ratio r=18r=\dfrac18. However, it doesn't start at k=0k=0, which is what we're used to.

  • We write out the first few terms of the series to figure out a convenient constant to factor out.

    ∑k=7∞18k=187+188+189+⋯=187(180+181+182+⋯ )=∑k=0∞187⋅18n\begin{align*}\sum_{k=7}^{\infty} \frac{1}{8^k}&=\frac{1}{8^7}+\frac{1}{8^8}+\frac{1}{8^9}+\cdots\\ &=\frac{1}{8^7}\left(\frac{1}{8^0}+\frac{1}{8^1}+\frac{1}{8^2}+\cdots\right)\\ &=\sum_{k=0}^\infty \frac{1}{8^7}\cdot \frac{1}{8^n}\end{align*}

    We now evaluate the series using Equation 3.2.2 in the CLP-2 text with r=18r=\dfrac18, a=187a=\dfrac{1}{8^7}.

    =187⋅11−18=17×86\begin{align*}&=\frac{1}{8^7}\cdot\dfrac{1}{1-\frac18} = \dfrac{1}{7\times 8^6}\end{align*}

  • Using the idea of Question 10, we express the series we're interested in as the difference of two series that we can easily evaluate.

    ∑k=7∞18k=∑k=0∞18k−∑k=0618k\begin{align*}\sum_{k=7}^{\infty} \frac{1}{8^k}&= \sum_{k=0}^{\infty} \frac{1}{8^k}- \sum_{k=0}^{6} \frac{1}{8^k}\end{align*}

    Using Equations 3.2.2 and 3.2.1 in the CLP-2 text,

    =11−18−1−1871−18=17×86\begin{align*}&=\frac{1}{1-\frac18}-\frac{1-\frac{1}{8^7}}{1-\frac18}\\ &=\frac{1}{7\times 8^6}\end{align*}
Q17Stage 2Past exam · 2016Q5

Show that the series ∑k=1∞(6k2−6(k+1)2)\displaystyle \sum_{k=1}^{\infty} \bigg( \frac{6}{k^2} - \frac{6}{(k+1)^2} \bigg) converges and find its limit.

Hint

When you see ∑k(⋯k⋯  −  ⋯k+1⋯)\displaystyle\sum_k \Big(\cdots \textcolor{red}{k}\cdots \ \ -\ \ \cdots \textcolor{red}{k+1}\cdots\Big), you should think “telescoping series."

Answer

66

Full solution

We recognize this as a telescoping series.

k6k2−6(k+1)2sk16−646−64264−696−69369−6166−6164616−6256−6255625−6366−6366636−6476−647⋮\begin{array}{c||cc||r} \mathbf{k}&\mathbf{\frac{6}{k^2}}&\mathbf{-\frac{6}{(k+1)^2}}&\mathbf{s_k}\\ \hline 1&6&\color{red}-\frac{6}{4}&6-\color{red}\frac64\\[7pt] 2&\color{red}\frac{6}{4}&\color{orange}-\frac{6}{9}&6-\color{orange}\frac69\\[7pt] 3&\color{orange}\frac{6}{9}&\color{green}-\frac{6}{16}&6-\color{green}\frac{6}{16}\\[7pt] 4&\color{green}\frac{6}{16}&\color{blue}-\frac{6}{25}&6-\color{blue}\frac{6}{25}\\[7pt] 5&\color{blue}\frac{6}{25}&\color{purple}-\frac{6}{36}&6-\color{purple}\frac{6}{36}\\[7pt] 6&\color{purple}\frac{6}{36}&-\frac{6}{47}&6-\frac{6}{47}\\[7pt] \vdots&&& \end{array}

When we compute the nthn^{\rm th} partial sum, i.e. the sum of of the first nn terms, successive terms cancel and only the first half of the first term, (6k2−6(k+1)2)∣k=1\Big( \frac{6}{k^2} - \frac{6}{(k+1)^2} \Big)\Big|_{k=1}, and the second half of the nthn^{\rm th} term, (6k2−6(k+1)2)∣k=n\Big( \frac{6}{k^2} - \frac{6}{(k+1)^2} \Big)\Big|_{k=n}, survive. That is:

sn=∑k=1n(6k2−6(k+1)2)=612−6(n+1)2\begin{align*} s_n = \sum_{k=1}^{n} \bigg( \frac{6}{k^2} - \frac{6}{(k+1)^2} \bigg) = \frac6{1^2} - \frac{6}{(n+1)^2} \end{align*}

Therefore, we can see directly that the sequence of partial sums {sn}\{ s_n \} is convergent:

lim⁡n→∞sn=lim⁡n→∞(6−6(n+1)2)=6\begin{align*} \lim_{n\to\infty} s_n = \lim_{n\to\infty} \bigg( 6- \frac{6}{(n+1)^2} \bigg) = 6 \end{align*}

By Definition 3.2.3 in the CLP-2 text the series is also convergent, with limit 66.

Q18Stage 2Past exam · 2016A

Find the sum of the convergent series ∑n=3∞( ⁣cos⁡(πn)−cos⁡(πn+1))\displaystyle\sum\limits_{n=3}^\infty \bigg( \!\cos\Big( \frac \pi n \Big) - \cos\Big( \frac \pi{n+1} \Big) \bigg).

Hint

When you see ∑n(⋯n⋯  −  ⋯n+1⋯)\displaystyle\sum_n \Big(\cdots \textcolor{red}{n}\cdots \ \ -\ \ \cdots \textcolor{red}{n+1}\cdots\Big), you should immediately think “telescoping series”. But be careful not to jump to conclusions — evaluate the nthn^{\rm th} partial sum explicitly.

Answer

cos⁡(π3)−cos⁡(0)=−12\displaystyle\cos\left( \frac \pi 3 \right) - \cos(0) = -\frac{1}{2}

Full solution

We recognize that this is a telescoping series, and set up a table to find the sequence of partial sums.

ncos⁡(πn)−cos⁡(πn+1)sn3cos⁡(π3)−cos⁡(π4)12−cos⁡(π4)4cos⁡(π4)−cos⁡(π5)12−cos⁡(π5)5cos⁡(π5)−cos⁡(π6)12−cos⁡(π6)6cos⁡(π6)−cos⁡(π7)12−cos⁡(π7)7cos⁡(π7)−cos⁡(π8)12−cos⁡(π8)8cos⁡(π8)−cos⁡(π9)12−cos⁡(π9)⋮\begin{array}{c||cc||r} \mathbf{n}&\mathbf{\cos\left(\frac{\pi}{n}\right)}&\mathbf{-\cos\left(\frac{\pi}{n+1}\right)}&\mathbf{s_n}\\[7pt] \hline 3&\cos\left(\frac{\pi}{3} \right)&\color{red}-\cos\left(\frac{\pi}{4}\right)&\frac12-\color{red}\cos\left(\frac{\pi}{4}\right)\\[7pt] 4&\color{red}\cos\left(\frac{\pi}{4}\right)&\color{orange}-\cos\left(\frac{\pi}{5}\right)&\frac12-\color{orange}\cos\left(\frac{\pi}{5}\right)\\[7pt] 5&\color{orange}\cos\left(\frac{\pi}{5}\right)&\color{green}-\cos\left(\frac{\pi}{6}\right)&\frac12-\color{green}\cos\left(\frac{\pi}{6}\right)\\[7pt] 6&\color{green}\cos\left(\frac{\pi}{6}\right)&\color{blue}-\cos\left(\frac{\pi}{7}\right)&\frac12-\color{blue}\cos\left(\frac{\pi}{7}\right)\\[7pt] 7&\color{blue}\cos\left(\frac{\pi}{7}\right)&\color{purple}-\cos\left(\frac{\pi}{8}\right)&\frac12-\color{purple}\cos\left(\frac{\pi}{8}\right)\\[7pt] 8&\color{purple}\cos\left(\frac{\pi}{8}\right)&-\cos\left(\frac{\pi}{9}\right)&\frac12-\cos\left(\frac{\pi}{9}\right)\\[7pt] \vdots&&& \end{array}

In the partial sum

sN=∑n=3N( ⁣cos⁡(πn)−cos⁡(πn+1))\begin{equation*} s_N = \sum_{n=3}^N \bigg( \!\cos\Big( \frac \pi n \Big) - \cos\Big( \frac \pi{n+1} \Big) \bigg) \end{equation*}

every term cancels except the first part of the first term (cos⁡(π3)=12\cos\big(\frac{\pi}{3}\big)=\tfrac12) and the second part of the last term (−cos⁡(πN+1)-\cos(\tfrac{\pi}{N+1})). So

sN=cos⁡(π3)−cos⁡(πN+1)=12−cos⁡(πN+1).\begin{align*}s_N &=\cos\Big( \frac \pi 3 \Big) - \cos\Big( \frac \pi{N+1} \Big)\\ &=\frac12 - \cos\Big( \frac \pi{N+1} \Big).\end{align*}

As N→∞N\to\infty, the argument πN+1\frac{\pi}{N+1} converges to 00, and cos⁡x\cos x is continuous at x=0x=0. By Definition 3.2.3 in the CLP-2 text, the value of the series is

lim⁡N→∞sN=lim⁡N→∞[12−cos⁡(πN+1))]=12−cos⁡(0)=−12\begin{align*}\lim_{N\to\infty} s_N &= \lim_{N\to\infty}\left[\frac12 - \cos\Big( \frac \pi{N+1} \Big) \bigg) \right]\\&= \frac12 - \cos(0) = -\frac{1}{2}\end{align*}
Q19Stage 2Past exam · 2016Q5

The nthn^{\rm th} partial sum of a series ∑n=1∞an\displaystyle \sum_{n=1}^{\infty}a_n is known to have the formula sn=1+3n5+4n\displaystyle s_n = \frac{1+3n}{5+4n}.

  1. Find an expression for ana_n, valid for n≥2n\ge2.

  2. Show that the series ∑n=1∞an\displaystyle \sum_{n=1}^{\infty}a_n converges and find its value.

Hint

Review Definition 3.2.3 in the CLP-2 text.

Answer

(a) an=1116n2+24n+5\displaystyle a_n= \frac{11}{16n^2 + 24n +5} (b) 34\dfrac{3}{4}

Full solution

(a) As in Question 2, since

sn−1=a1+a2+⋯+an−1sn=a1+a2+⋯+an−1+an\begin{align*} s_{n-1}& = a_1+a_2+ \cdots + a_{n-1} \\ s_{n}& = a_1+a_2+ \cdots + a_{n-1} +a_n\\ \end{align*}

we can find ana_n by subtracting:

an=sn−sn−1=1+3n5+4n−1+3(n−1)5+4(n−1)=3n+14n+5−3n−24n+1=(3n+1)(4n+1)−(3n−2)(4n+5)(4n+1)(4n+5)=1116n2+24n+5\begin{align*} a_n &= s_n - s_{n-1} \\ &=\frac{1 + 3n}{5+4n} - \frac{1 + 3(n-1)}{5+4(n-1)} = \frac{3n+1}{4n+5} - \frac{3n - 2}{4n+1}\\ &= \frac{(3n+1)(4n+1) - (3n - 2) (4n+5)}{(4n+1)(4n+5)} \\ &= \frac{11}{16n^2 + 24n +5} \end{align*}

(b) Using Definition 3.2.3 in the CLP-2 text,

∑n=1∞an=lim⁡n→∞sn=lim⁡n→∞1+3n5+4n=lim⁡n→∞1/n+35/n+4=0+30+4=34\begin{align*} \sum_{n=1}^\infty a_n&= \lim_{n \to \infty} s_n = \lim_{n \to \infty} \frac{1 + 3n}{5+4n} = \lim_{n \to \infty} \frac{1/n + 3}{5/n+4} = \frac{0+3}{0+4} = \frac{3}{4} \end{align*}

The series converges to 34\dfrac{3}{4}.

Q20Stage 2Past exam · 2013A

Find the sum of the series ∑n=2∞3⋅4n+18⋅5n\displaystyle\sum\limits_{n=2}^\infty \frac{3\cdot 4^{n+1}}{8\cdot 5^n}. Simplify your answer completely.

Hint

This is a special case of a general series whose sum we know.

Answer

245\dfrac{24}{5}

Full solution

What we have is a geometric series, but we need to get it into the proper form before we can evaluate it.

∑n=2∞3⋅4n+18⋅5n=∑n=2∞3⋅4⋅4n8⋅5n=32∑n=2∞(45)n\begin{align*} \sum_{n=2}^\infty \frac{3\cdot 4^{n+1}}{8\cdot 5^n} &= \sum_{n=2}^\infty \frac{3\cdot4\cdot 4^{n}}{8\cdot 5^n}= \frac{3}{2}\sum_{n=2}^\infty \Big(\frac{4}{5}\Big)^n \end{align*}

  • If we factor our (45)2\left(\frac45\right)^2, we can change our index to something more convenient.

    32∑n=2∞(45)n=32∑n=2∞(45)2(45)n−2=32∑n=0∞(45)2(45)n\begin{align*}\frac{3}{2}\sum_{n=2}^\infty \left(\frac{4}{5}\right)^n&= \frac{3}{2}\sum_{n=2}^\infty \left(\frac{4}{5}\right)^2\left(\frac45\right)^{n-2}\\ &=\frac{3}{2}\sum_{n=0}^\infty \left(\frac{4}{5}\right)^2\left(\frac45\right)^{n}\end{align*}

    We use Equation 3.2.2 in the CLP-2 text with r=45r=\dfrac45.

    =32(45)2⋅11−45=245\begin{align*}&=\frac{3}{2}\left(\frac{4}{5}\right)^2\cdot\frac{1}{1-\frac45}=\frac{24}{5}\end{align*}

  • Using the idea of Question 10, we view our series as a more convenient series, minus a few initial terms.

    32∑n=2∞(45)n=32([∑n=0∞(45)n]−(45)1−(45)0)=32(∑n=0∞(45)n−95)\begin{align*}\frac{3}{2}\sum_{n=2}^\infty \left(\frac{4}{5}\right)^n&= \frac{3}{2}\left(\left[\sum_{n=0}^\infty \left(\frac{4}{5}\right)^n\right] - \left(\frac{4}{5}\right)^1- \left(\frac{4}{5}\right)^0\right)\\&= \frac{3}{2}\left(\sum_{n=0}^\infty \left(\frac{4}{5}\right)^n - \frac{9}{5}\right)\end{align*}

    We use Equation 3.2.2 in the CLP-2 text with r=45r=\dfrac45.

    =32(11−45−95)=245\begin{align*}&=\frac{3}{2}\left(\frac{1}{1-\frac45} - \frac{9}{5}\right)=\frac{24}{5}\end{align*}
Q21Stage 2Past exam · 2016Q5

Relate the number 0.23ˉ=0.233333…0.2\bar{3} = 0.233333\ldots to the sum of a geometric series, and use that to represent it as a rational number (a fraction or combination of fractions, with no decimals).

Hint

Review Example 3.2.5 in the CLP-2 text. To write the number as a geometric series, the first few terms might not fit the pattern of the rest of the terms.

Answer

730\dfrac{7}{30}

Full solution

The number is:

0.2+3100+31000+310000+⋯=15+3102+3103+3104+⋯=15+3102(1100+1101+1102+⋯ )=15+3102∑n=0∞110n\begin{align*}0.2 + \frac{3}{100} + \frac{3}{1000} + \frac{3}{10000} + \cdots &=\frac{1}{5}+\frac{3}{10^2}+\frac{3}{10^3}+\frac{3}{10^4}+\cdots\\ &=\frac{1}{5}+\frac{3}{10^2}\left(\frac{1}{10^0}+\frac{1}{10^1}+\frac{1}{10^2}+\cdots\right)\\ & =\frac15 + \frac{3}{10^2}\sum_{n=0}^{\infty} \frac{1}{10^{n}}\end{align*}

We use Equation 3.2.2 in the CLP-2 text with r=110r=\dfrac{1}{10}.

=15+3102⋅11−110=15+130=730\begin{align*}&=\frac{1}{5}+\frac{3}{10^2}\cdot\frac{1}{1-\frac{1}{10}}\\ &=\frac15+\frac{1}{30} = \frac{7}{30}\end{align*}
Q22Stage 2Past exam · 2012A

Express 2.656565…2.656565\ldots as a rational number, i.e. in the form p/qp/q where pp and qq are integers.

Hint

Start by writing it as a geometric series.

Answer

26399\dfrac{263}{99}

Full solution

The number is:

2+65100+6510000+651000000+⋯=2+65100+651002+651003+⋯=2+65100∑n=0∞1100n\begin{align*}2+\frac{65}{100}+\frac{65}{10000}+\frac{65}{1000000}+\cdots&= 2 + \frac{65}{100} + \frac{65}{100^2} + \frac{65}{100^3} + \cdots \\ &= 2 + \frac{65}{100}\sum\limits_{n=0}^{\infty} \frac{1}{100^n}\end{align*}

We use Equation 3.2.2 in the CLP-2 text with r=1100r=\dfrac{1}{100}.

=2+65100⋅11−1100=2+6599=26399\begin{align*}&= 2 + \frac{65}{100}\cdot\frac{1}{1-\frac{1}{100}}\\ &=2+\frac{65}{99} = \frac{263}{99}\end{align*}
Q23Stage 2Past exam · 2016A

Express the decimal 0.321‾=0.321321321…0.\overline{321}=0.321321321\ldots as a fraction.

Hint

Review Example 3.2.5 in the CLP-2 text. Since the pattern repeats every three decimals, your common ratio rr will be 1103\dfrac{1}{10^3}.

Answer

321999=107333\dfrac{321}{999}= \dfrac{107}{333}

Full solution

The number is:

0.321‾=0.321321321…=3211000+321106+321109+⋯=3211000((1103)0+(1103)1+(1103)2+⋯ )=3211000∑n=0∞(1103)n\begin{align*}0.\overline{321}&=0.321321321\ldots\\ &= \frac{321}{1000} + \frac{321}{10^6} + \frac{321}{10^9} + \cdots \\ &= \frac{321}{1000} \left(\left(\frac{1}{10^3}\right)^0+ \left(\frac{1}{10^3}\right)^1 + \left(\frac{1}{10^3}\right)^2 + \cdots \right)\\ &= \frac{321}{1000} \sum_{n=0}^{\infty} \left(\frac{1}{10^3}\right)^n\end{align*}

We use Equation 3.2.2 in the CLP-2 text with r=1103r=\dfrac{1}{10^3}.

=3211000⋅11−1103=321999=107333\begin{align*}&=\frac{321}{1000} \cdot\frac{1}{1-\frac{1}{10^3}}=\frac{321}{999}=\frac{107}{333}\end{align*}
Q24Stage 2Past exam · 2015A

Find the value of the convergent series

∑n=2∞(2n+13n+12n−1−12n+1)\begin{align*} \sum_{n=2}^\infty \bigg( \frac{2^{n+1}}{3^n} + \frac1{2n-1} - \frac1{2n+1} \bigg) \end{align*}

Simplify your answer completely.

Hint

Split the series into two parts.

Answer

33

Full solution

We split the sum into two parts.

∑n=2∞(2n+13n+12n−1−12n+1)=∑n=2∞2n+13n+∑n=2∞(12n−1−12n+1)\begin{align*} \sum_{n=2}^\infty \bigg( \frac{2^{n+1}}{3^n} + \frac1{2n-1} - \frac1{2n+1} \bigg)&= \color{blue} \sum_{n=2}^\infty \frac{2^{n+1}}{3^n} \quad \textcolor{black}{+}\quad \color{red} \sum_{n=2}^\infty \bigg( \frac1{2n-1} - \frac1{2n+1} \bigg) \end{align*}

The first part is a geometric series.

∑n=2∞2n+13n=∑n=0∞2n+33n+2=∑n=0∞2332⋅(23)n\begin{align*}\color{blue} \sum_{n=2}^\infty \frac{2^{n+1}}{3^n} &= \sum_{n=0}^\infty \frac{2^{n+3}}{3^{n+2}} = \sum_{n=0}^\infty \frac{2^3}{3^2}\cdot\left( \frac{2}{3}\right)^n\end{align*}

We use Equation 3.2.2 in the CLP-2 text with r=23r=\dfrac{2}{3} and a=89a=\dfrac{8}{9}.

=89⋅11−23=83\begin{align*}&=\frac{8}{9}\cdot\frac{1}{1-\frac23}=\color{blue}\frac{8}{3}\end{align*}

The second part is a telescoping series. Let's make a table to see how it cancels.

n12n−1−12n+1sn213−1513−15315−1713−17417−1913−19519−11113−1116111−11313−1137113−11513−115⋮\begin{array}{c||cc||r} \mathbf{n}&\mathbf{\frac{1}{2n-1}}&\mathbf{-\frac{1}{2n+1}}&\mathbf{s_n}\\[7pt] \hline 2&\frac{1}{3}&\color{red}-\frac{1}{5}&\frac{1}{3}-\color{red}\frac{1}{5}\\[7pt] 3&\color{red}\frac{1}{5}&\color{orange}-\frac{1}{7}&\frac{1}{3}-\color{orange}\frac{1}{7}\\[7pt] 4&\color{orange}\frac{1}{7}&\color{green}-\frac{1}{9}&\frac{1}{3}-\color{green}\frac{1}{9}\\[7pt] 5&\color{green}\frac{1}{9}&\color{blue}-\frac{1}{11}&\frac{1}{3}-\color{blue}\frac{1}{11}\\[7pt] 6&\color{blue}\frac{1}{11}&\color{purple}-\frac{1}{13}&\frac{1}{3}-\color{purple}\frac{1}{13}\\[7pt] 7&\color{purple}\frac{1}{13}&\color{black}-\frac{1}{15}&\frac{1}{3}-\color{black}\frac{1}{15}\\[7pt] \vdots&&& \end{array}

After adding terms n=2n=2 through n=Nn=N, the partial sum is

sN=13−12N+1\begin{align*}s_N&=\frac{1}{3}-\frac{1}{2N+1}\end{align*}

because all the terms except the first part of the n=2n=2 term, and the last part of the n=Nn=N term, cancel. Then:

∑n=2∞(12n−1−12n+1)=lim⁡N→∞sN=lim⁡N→∞13−12N+1=13\begin{align*}\color{red}\sum_{n=2}^\infty \bigg(\frac1{2n-1} - \frac1{2n+1}\bigg) &=\lim_{N \to \infty}s_N = \lim_{N \to \infty}\frac{1}{3}-\frac{1}{2N+1}\\ &=\color{red}\frac{1}{3}\end{align*}

All together,

∑n=2∞(2n+13n+12n−1−12n+1)=∑n=2∞2n+13n+∑n=2∞(12n−1−12n+1)=83+13=3\begin{align*}\sum_{n=2}^\infty \bigg( \frac{2^{n+1}}{3^n} + \frac1{2n-1} - \frac1{2n+1} \bigg)&= \color{blue} \sum_{n=2}^\infty \frac{2^{n+1}}{3^n} \quad \textcolor{black}{+}\quad \color{red} \sum_{n=2}^\infty \bigg( \frac1{2n-1} - \frac1{2n+1} \bigg)\\ &=\textcolor{blue}{\frac83}+\textcolor{red}{\frac13}=3\end{align*}
Q25Stage 2Past exam · M105 2014A

Evaluate

∑n=1∞[(13)n+(−25)n−1]\begin{align*} \sum_{n=1}^\infty\bigg[{\Big(\frac{1}{3}\Big)}^n + {\Big(-\frac{2}{5}\Big)}^{n-1}\bigg] \end{align*}
Hint

Split the series into two parts.

Answer

12+57=1714\dfrac{1}{2}+\dfrac{5}{7} =\dfrac{17}{14}

Full solution

We split the sum into two parts.

∑n=1∞[(13)n+(−25)n−1]=∑n=1∞(13)n+∑n=1∞(−25)n−1\begin{align*}\sum_{n=1}^\infty\bigg[{\Big(\frac{1}{3}\Big)}^n + {\Big(-\frac{2}{5}\Big)}^{n-1}\bigg] &= \sum_{n=1}^\infty {\Big(\frac{1}{3}\Big)}^n + \sum_{n=1}^\infty {\Big(-\frac{2}{5}\Big)}^{n-1}\end{align*}

Both are geometric series.

=∑n=0∞(13)n+1+∑n=0∞(−25)n=13∑n=0∞(13)n+∑n=0∞(−25)n\begin{align*}&= \sum_{n=0}^\infty {\Big(\frac{1}{3}\Big)}^{n+1} + \sum_{n=0}^\infty {\Big(-\frac{2}{5}\Big)}^{n} \\&= \frac{1}{3}\sum_{n=0}^\infty {\Big(\frac{1}{3}\Big)}^{n} + \sum_{n=0}^\infty {\Big(-\frac{2}{5}\Big)}^{n}\end{align*}

We use Equation 3.2.2 in the CLP-2 text with a1=13a_1=\dfrac{1}{3} and r1=13r_1=\dfrac{1}{3}, then with a2=1a_2=1 and r2=−25r_2=-\dfrac{2}{5}.

=13⋅11−13+11+25=12+57=1714\begin{align*}&=\frac{1}{3}\cdot\frac{1}{1-\frac13}+\frac{1}{1+\frac25}\\ &=\frac{1}{2}+\frac{5}{7}=\frac{17}{14}\end{align*}
Q26Stage 2Past exam · 2014D

Find the sum of the series ∑n=0∞1+3n+14n\displaystyle\sum_{n=0}^\infty\frac{1+3^{n+1}}{4^n}.

Hint

Split the series into two parts.

Answer

403\dfrac{40}{3}

Full solution

We split the sum into two parts.

∑n=0∞1+3n+14n=∑n=0∞14n+∑n=0∞3n+14n=∑n=0∞14n+3∑n=0∞(34)n\begin{align*}\sum_{n=0}^\infty\frac{1+3^{n+1}}{4^n} &= \sum_{n=0}^\infty \frac{1}{4^n} + \sum_{n=0}^\infty \frac{3^{n+1}}{4^n} \\&= \sum_{n=0}^\infty \frac{1}{4^n} +3 \sum_{n=0}^\infty\left( \frac{3}{4}\right)^n\end{align*}

Using Equation 3.2.2 in the CLP-2 text,

=11−14+31−34=43+12=403\begin{align*}&=\frac{1}{1-\frac14}+\frac{3}{1-\frac{3}{4}}\\ &=\frac{4}{3}+12=\frac{40}{3}\end{align*}
Q27Stage 2

Evaluate ∑n=5∞log⁡(n−3n)\displaystyle\sum_{n=5}^\infty \log\left(\frac{n-3}{n}\right).

Hint

Use logarithm rules to turn this into a more obvious telescoping series.

Answer

The series diverges to −∞-\infty.

Full solution

Using logarithm rules, we see

∑n=5∞log⁡(n−3n)=∑n=5∞[log⁡(n−3)−log⁡n]\displaystyle\sum_{n=5}^\infty \log\left(\frac{n-3}{n}\right)= \displaystyle\sum_{n=5}^\infty \big[\log(n-3)-\log n\big]

which looks like a telescoping series. Let's make a table to figure out the partial sums.

nlog⁡(n−3)−log⁡nsn5log⁡2−log⁡5log⁡2−log⁡56log⁡3−log⁡6log⁡2+log⁡3−log⁡5−log⁡67log⁡4−log⁡7log⁡2+log⁡3+log⁡4−log⁡5−log⁡6−log⁡78log⁡5−log⁡8log⁡2+log⁡3+log⁡4−log⁡6−log⁡7−log⁡89log⁡6−log⁡9log⁡2+log⁡3+log⁡4−log⁡7−log⁡8−log⁡910log⁡7−log⁡10log⁡2+log⁡3+log⁡4−log⁡8−log⁡9−log⁡1011log⁡8−log⁡11log⁡2+log⁡3+log⁡4−log⁡9−log⁡10−log⁡11⋮\begin{array}{c||cc||c} \mathbf{n}&\mathbf{\log(n-3)}&\mathbf{-\log n}&\mathbf{s_n}\\[7pt] \hline 5&\log 2&\color{red}-\log 5&\log 2 - \color{red}\log 5\\[7pt] 6&\log 3&\color{green}-\log 6&\log 2 +\log 3- \color{red}\log 5\color{green}-\log 6\\[7pt] 7&\log 4&\color{purple}-\log 7&\log 2 +\log 3+\log 4- \color{red}\log 5\color{green}-\log 6 \color{purple}-\log 7\\[7pt] 8&\color{red}\log 5&\color{blue}-\log 8&\log 2 +\log 3+\log 4 \color{green}-\log 6 \color{purple}-\log 7\color{blue}-\log 8\\[7pt] 9&\color{green}\log 6&\color{orange}-\log 9&\log 2 +\log 3+\log 4 \color{purple}-\log 7\color{blue}-\log 8\color{orange}-\log 9\\[7pt] 10&\color{purple}\log 7&\color{brown}-\log 10&\log 2 +\log 3+\log 4 \color{blue}-\log 8\color{orange}-\log 9\color{brown}-\log 10\\[7pt] 11&\color{blue}\log 8&\color{yellow}-\log 11&\log 2 +\log 3+\log 4 \color{orange}-\log 9\color{brown}-\log 10\color{yellow}-\log 11\\[7pt] \vdots&&& \end{array}

There is a “lag" before the terms cancel, which is why they “build up" more than we saw in past examples. Still, we can clearly see the NNth partial sum:

∑n=5N(log⁡(n−3)−log⁡(n))=log⁡2+log⁡3+log⁡4−log⁡(N−2)−log⁡(N−1)−log⁡(N)=log⁡(24N(N−1)(N−2))\begin{align*}\sum_{n=5}^N \Big(\log(n-3)-\log(n)\Big)&=\log 2+\log 3 + \log 4 - \log(N-2)-\log(N-1)-\log(N)\\ &=\log \left(\frac{24}{N(N-1)(N-2)}\right)\end{align*}

when N≥7N \geq 7. So,

∑n=5∞(log⁡(n−3)−log⁡(n))=lim⁡N→∞sN=lim⁡N→∞log⁡(24N(N−1)(N−2))=−∞\begin{align*}\sum_{n=5}^\infty \Big(\log(n-3)-\log(n)\Big)&= \lim_{N \to \infty}s_N\\ &=\lim_{N \to \infty}\log \left(\frac{24}{N(N-1)(N-2)}\right) \\&=-\infty\end{align*}
Q28Stage 2

Evaluate ∑n=2∞(2n−1n+1−1n−1)\displaystyle\sum_{n=2}^\infty \left(\frac{2}{n}-\frac{1}{n+1}-\frac{1}{n-1} \right).

Hint

This is a telescoping series.

Answer

−12-\dfrac12

Full solution

This is a telescoping series. Let's investigate it in the usual way. To make the pattern of cancellation clearer, we express 2n−1n+1−1n−1=1n+1n−1n+1−1n−1\frac{2}{n}-\frac{1}{n+1}-\frac{1}{n-1}= \frac{1}{n}+\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n-1}, and leave the fractions in the middle of the table unsimplified. Then every fraction has numerator one and two terms with the same denominator and opposite sign cancel.

n1n1n−1n+1−1n−1sn21212−13−1112+12−13−1131313−14−1212+13−14−1141414−15−1312+14−15−1151515−16−1412+15−16−1161616−17−1512+16−17−1171717−18−1612+17−18−1181818−19−1712+18−19−11⋮\begin{array}{c||cccc||c} \mathbf{n}&\mathbf{\dfrac{1}{n}}&\mathbf{\dfrac{1}{n}}&\mathbf{-\dfrac{1}{n+1}}&\mathbf{-\dfrac{1}{n-1}}&\mathbf{s_n}\\[7pt] \hline 2&\dfrac{1}{2}&{\color{green}{\dfrac{1}{2}}}&{\color{red}{-\dfrac{1}{3}}}&-\dfrac{1}{1}& \dfrac{1}{2}+{\color{green}{\dfrac{1}{2}}} {\color{red}{-\dfrac{1}{3}}}-\dfrac{1}{1}\\[10pt] 3&{\color{red}{\dfrac{1}{3}}}&{\color{blue}{\dfrac{1}{3}}}&{\color{orange}{-\dfrac{1}{4}}}&{\color{green}{-\dfrac{1}{2}}}& \dfrac{1}{2}+{\color{blue}{\dfrac{1}{3}}}{\color{orange}{-\dfrac{1}{4}}}-\dfrac{1}{1}\\[10pt] 4&{\color{orange}{\dfrac{1}{4}}}&{\color{green}{\dfrac{1}{4}}}&{\color{red}{-\dfrac{1}{5}}}&{\color{blue}{-\dfrac{1}{3}}}& \dfrac{1}{2}+{\color{green}{\dfrac{1}{4}}} {\color{red}{-\dfrac{1}{5}}}-\dfrac{1}{1}\\[10pt] 5&{\color{red}{\dfrac{1}{5}}}&{\color{blue}{\dfrac{1}{5}}}&{\color{orange}{-\dfrac{1}{6}}}&{\color{green}{-\dfrac{1}{4}}}& \dfrac{1}{2}+{\color{blue}{\dfrac{1}{5}}}{\color{orange}{-\dfrac{1}{6}}}-\dfrac{1}{1}\\[10pt] 6&{\color{orange}{\dfrac{1}{6}}}&{\color{green}{\dfrac{1}{6}}}&{\color{red}{-\dfrac{1}{7}}}&{\color{blue}{-\dfrac{1}{5}}}& \dfrac{1}{2}+{\color{green}{\dfrac{1}{6}}}{\color{red}{-\dfrac{1}{7}}}-\dfrac{1}{1}\\[10pt] 7&{\color{red}{\dfrac{1}{7}}}&{\color{blue}{\dfrac{1}{7}}}&{\color{orange}{-\dfrac{1}{8}}}&{\color{green}{-\dfrac{1}{6}}}& \dfrac{1}{2}+{\color{blue}{\dfrac{1}{7}}} {\color{orange}{-\dfrac{1}{8}}}-\dfrac{1}{1}\\[10pt] 8&{\color{orange}{\dfrac{1}{8}}}&\dfrac{1}{8}&-\dfrac{1}{9}&{\color{blue}{-\dfrac{1}{7}}}& \dfrac{1}{2}+\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{1}\\[10pt] \vdots&&&& \end{array}

Concentrate on any row nn, except the very first row and the very last row. The first 1n\frac{1}{n} in that row cancels the −1n-\frac{1}{n} in the middle of the row above it, and the second 1n\frac{1}{n} in that row cancels the −1n-\frac{1}{n} at the end of the row below it. As far as the first (n=2n=2) row is concerned, the first 12\frac{1}{2} and the last −11-\frac{1}{1} never get cancelled out because there is no row above the first one. And as far as the very last row is concerned, the two middle terms never get cancelled out because there is no row after the last one. So the partial sum is

sN=∑n=2N(2n−1n+1−1n−1)=12−11⏞from the first row+1N−1N+1⏞from the last row\begin{align*} s_N&=\sum_{n=2}^N \left(\frac{2}{n}-\frac{1}{n+1}-\frac{1}{n-1} \right) =\overbrace{\frac{1}{2}-\frac{1}{1}}^{\text{from the first row}} +\overbrace{\frac{1}{N}-\frac{1}{N+1}}^{\text{from the last row}} \end{align*}

There is another purely algebraic way to find the same sNs_N, motivated by the above discussion.

sN=∑n=2N(2n−1n+1−1n−1)=∑n=2N(1n−1n+1)+∑n=2N(1n−1n−1)\begin{align*} s_N&=\sum_{n=2}^N \left(\frac{2}{n}-\frac{1}{n+1}-\frac{1}{n-1} \right) \\ &=\sum_{n=2}^N \left(\frac{1}{n}-\frac{1}{n+1}\right) +\sum_{n=2}^N\left(\frac{1}{n}-\frac{1}{n-1} \right) \end{align*}

The first half

∑n=2N(1n−1n+1)=(12−13)+(13−14)+(14−15)+⋯+(1N−1N+1)=[12−1N+1]\begin{align*} \sum_{n=2}^N \left(\frac{1}{n}-\frac{1}{n+1}\right) &=\left(\frac{1}{2}{\color{red}{-\frac{1}{3}}}\right) + \left({\color{red}{\frac{1}{3}}} {\color{blue}{-\frac{1}{4}}}\right) + \left({\color{blue}{\frac{1}{4}}}{\color{red}{-\frac{1}{5}}}\right) + \cdots +\left({\color{red}{\frac{1}{N}}}-\frac{1}{N+1}\right) \\ &=\left[\frac{1}{2}-\frac{1}{N+1}\right] \end{align*}

and the second half

∑n=2N(1n−1n−1)=(12−11)+(13−12)+(14−13)+⋯+(1N−1N−1)={−11+1N}\begin{align*} \sum_{n=2}^N\left(\frac{1}{n}-\frac{1}{n-1} \right) &=\left({\color{red}{\frac{1}{2}}}-\frac{1}{1} \right) + \left({\color{blue}{\frac{1}{3}}}{\color{red}{-\frac{1}{2}}} \right) + \left({\color{red}{\frac{1}{4}}}{\color{blue}{-\frac{1}{3}}}\right) +\cdots + \left(\frac{1}{N}-{\color{red}{\frac{1}{N-1}}} \right) \\ &=\left\{-\frac{1}{1}+\frac{1}{N}\right\} \end{align*}

So

sN=[12−1N+1]+{−11+1N}=−12+1N−1N+1\begin{align*} s_N&=\left[\frac{1}{2}-\frac{1}{N+1}\right]+\left\{-\frac{1}{1}+\frac{1}{N}\right\} \\ &=-\frac{1}{2}+\frac{1}{N}-\frac{1}{N+1} \end{align*}

and the limit

∑n=2∞(2n−1n+1−1n−1)=lim⁡N→∞sN=lim⁡N→∞[−12+1N−1N+1]=−12\begin{align*} \sum_{n=2}^\infty \left(\frac{2}{n}-\frac{1}{n+1}-\frac{1}{n-1} \right)&=\lim_{N \to \infty}s_N\\&=\lim_{N \to \infty} \left[-\frac{1}{2} +\frac{1}{N}-\frac{1}{N+1}\right]\\ &=-\frac{1}{2} \end{align*}

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q29Stage 3

An infinitely long, flat cliff has stones hanging off it, attached to thin wire of negligible mass. Starting at position x=1x=1, every metre (at position xx, where xx is some whole number) the stone has mass 14x\dfrac{1}{4^x} kg and is hanging 2x2^x metres below the top of the cliff.

Figure from prob_s3.2, line 1

Figure from prob_s3.2, line 1

Hint

The stone at position xx has mass 14x\dfrac{1}{4^x} kg, and we have to pull it a distance of 2x2^x metres. From this, you can find the work involved in pulling up a single stone. Then, add up the work involved in pulling up all the stones.

Answer

9.8 J

Full solution

The stone at position xx has mass 14x\dfrac{1}{4^x} kg, and we have to pull it a distance of 2x2^x metres, so the work involved in moving that one stone is

(14x kg)(2x m)(9.8 msec2)=9.82x J\left(\frac{1}{4^x}\text{ kg}\right)\left(2^x\text{ m}\right)\left(9.8 ~\frac{\text{m}}{\text{sec}^2}\right) = \frac{9.8}{2^x}\text{ J}

Therefore, the work to move all the stones is:

∑x=1∞9.82x=∑x=0∞9.82x+1=∑x=0∞9.82⋅12x=9.82⋅11−12=9.8 J\begin{align*} \sum_{x=1}^{\infty}\frac{9.8}{2^x}& =\sum_{x=0}^{\infty}\frac{9.8}{2^{x+1}}\\ &=\sum_{x=0}^{\infty}\frac{9.8}{2}\cdot\frac{1}{2^{x}} =\frac{9.8}{2}\cdot\frac{1}{1-\frac12}=9.8 \text{ J} \end{align*}
Q30Stage 3

Find the combined volume of an infinite collection of spheres, where for each whole number n=1, 2, 3, …n=1,\,2,\,3,\,\ldots there is exactly one sphere of radius 1πn\dfrac{1}{\pi^n}.

Hint

The volume of a sphere of radius rr is 43πr3\dfrac{4}{3}\pi r^3.

Answer

4π3(π3−1)\dfrac{4\pi}{3\left(\pi^3-1\right)}

Full solution

The volume of a sphere of radius 1πn\dfrac{1}{\pi^n} is

vn=43π(1πn)3=4π3(1π3)n\begin{align*}v_n&=\dfrac{4}{3}\pi \left(\dfrac{1}{\pi^n}\right)^3=\frac{4\pi}{3}\left(\frac{1}{\pi^3}\right)^{n}\end{align*}

So, the volume of all the spheres together is:

∑n=1∞vn=∑n=1∞4π3(1π3)n=∑n=0∞4π3(1π3)n+1=∑n=0∞43π2(1π3)n\begin{align*}\sum_{n=1}^\infty v_n&=\sum_{n=1}^\infty \frac{4\pi}{3}\left(\frac{1}{\pi^3}\right)^{n}\\ &=\sum_{n=0}^\infty \frac{4\pi}{3}\left(\frac{1}{\pi^3}\right)^{n+1}\\ &=\sum_{n=0}^\infty \frac{4}{3\pi^2}\left(\frac{1}{\pi^3}\right)^{n}\end{align*}

We use Equation 3.2.2 in the CLP-2 text with a=43π2a= \dfrac{4}{3\pi^2} and r=1π3r=\dfrac{1}{\pi^3}.

=43π2⋅11−1π3=4π3(π3−1)\begin{align*}&= \dfrac{4}{3\pi^2}\cdot\frac{1}{1-\frac{1}{\pi^3}} = \frac{4\pi}{3\left(\pi^3-1\right)}\end{align*}
Q31Stage 3

Evaluate ∑n=3∞(sin⁡2n2n+cos⁡2(n+1)2n+1)\displaystyle\sum_{n=3}^\infty \left(\frac{\sin^2 n}{2^n}+\frac{\cos^2(n+1)}{2^{n+1}} \right).

Hint

Use the properties of a telescoping series to simplify the terms.
Recall sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1.

Answer

sin⁡238+32≈32.0025\displaystyle \frac{\sin^23}{8}+32\approx 32.0025

Full solution

Let's make a table. Keep in mind cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1.

nsin⁡2n2ncos⁡2(n+1)2n+1sn3sin⁡2323cos⁡2424sin⁡2323+cos⁡24244sin⁡2424cos⁡2525sin⁡2323+124+cos⁡25255sin⁡2525cos⁡2626sin⁡2323+124+125+cos⁡26266sin⁡2626cos⁡2727sin⁡2323+124+125+126+cos⁡27277sin⁡2727cos⁡2828sin⁡2323+124+125+126+127+cos⁡2828⋮\begin{array}{c||cc||c} \mathbf{n}&\mathbf{\dfrac{\sin^2n}{2^n}}&\mathbf{\dfrac{\cos^2(n+1)}{2^{n+1}}}&\mathbf{s_n}\\[10pt] \hline 3&\dfrac{\sin^23}{2^3}&\color{red}\dfrac{\cos^24}{2^4}& \dfrac{\sin^23}{2^3}+\color{red}\dfrac{\cos^24}{2^4}\\[10pt] 4&\color{red}\dfrac{\sin^24}{2^4}&\color{green}\dfrac{\cos^25}{2^5}& \dfrac{\sin^23}{2^3}+{\color{red}\dfrac{1}{2^4}}+\color{green}\dfrac{\cos^25}{2^5}\\[10pt] 5&\color{green}\dfrac{\sin^25}{2^5}&\color{blue}\dfrac{\cos^26}{2^6}& \dfrac{\sin^23}{2^3}+{\color{red}\dfrac{1}{2^4}}+{\color{green}\dfrac{1}{2^5}}+\color{blue}\dfrac{\cos^26}{2^6}\\[10pt] 6&\color{blue}\dfrac{\sin^26}{2^6}&\color{orange}\dfrac{\cos^27}{2^7}& \dfrac{\sin^23}{2^3}+{\color{red}\dfrac{1}{2^4}}+{\color{green}\dfrac{1}{2^5}}+{\color{blue}\dfrac{1}{2^6}}+\color{orange}\dfrac{\cos^27}{2^7}\\[10pt] 7&\color{orange}\dfrac{\sin^27}{2^7}&\color{brown}\dfrac{\cos^28}{2^8}& \dfrac{\sin^23}{2^3}+{\color{red}\dfrac{1}{2^4}}+{\color{green}\dfrac{1}{2^5}}+ {\color{blue}\dfrac{1}{2^6}}+{\color{orange}\dfrac{1}{2^7}}+\color{brown}\dfrac{\cos^28}{2^8}\\[10pt] \vdots&&& \end{array}

This gives us an equation for the partial sum sNs_N, when N≥4N \geq 4:

sN=∑n=3N(sin⁡2n2n+cos⁡2(n+1)2n+1)=sin⁡2323+(∑n=4N12n)+cos⁡2(N+1)2N+1\begin{align*}s_N&=\sum_{n=3}^N\left(\frac{\sin^2 n}{2^n}+\frac{\cos^2(n+1)}{2^{n+1}} \right)\\ &=\frac{\sin^23}{2^3} +\left( \sum_{n=4}^{N}\frac{1}{2^n} \right)+\frac{\cos^2(N+1)}{2^{N+1}}\end{align*}

Using Definition 3.2.3 in the CLP-2 text, our series evaluates to:

lim⁡N→∞sN=lim⁡N→∞[sin⁡2323+(∑n=4N12n)+cos⁡2(N+1)2N+1]=sin⁡238+[lim⁡N→∞cos⁡2(N+1)2N+1]+∑n=4∞12n\begin{align*}\lim_{N \to \infty}s_N &=\lim_{N \to \infty}\left[\frac{\sin^23}{2^3} +\left( \sum_{n=4}^{N}\frac{1}{2^n} \right)+\frac{\cos^2(N+1)}{2^{N+1}}\right]\\ &=\frac{\sin^23}{8}+\left[\lim_{N \to \infty}\frac{\cos^2(N+1)}{2^{N+1}}\right] +\sum_{n=4}^\infty \frac{1}{2^n}\end{align*}

We evaluate the limit using the squeeze theorem; the series is geometric.

=sin⁡238+0+∑n=0∞12n+4=sin⁡238+124∑n=0∞12n\begin{align*}&=\frac{\sin^23}{8}+0+\sum_{n=0}^\infty \frac{1}{2^{n+4}} \\&=\frac{\sin^23}{8}+\frac{1}{2^4}\sum_{n=0}^\infty \frac{1}{2^{n}}\end{align*}

Using Equation 3.2.2 in the CLP-2 text,

=sin⁡238+12411−12=sin⁡238+18≈0.1275\begin{align*}&=\frac{\sin^23}{8}+\frac{1}{2^4}\frac{1}{1-\frac12}\\ &=\frac{\sin^23}{8}+\frac{1}{8}\approx 0.1275\end{align*}
Q32Stage 3

Suppose a series ∑n=1∞an\displaystyle\sum_{n=1}^\infty a_n has sequence of partial sums {SN}\{S_N\}, and the series ∑N=1∞SN\displaystyle\sum_{N=1}^\infty S_N has sequence of partial sums {SM}={∑N=1MSN}\{\mathscr S_M \}=\left\{ \displaystyle\sum_{N=1}^M S_N\right\}.

If SM=M+1M\mathscr S_M = \dfrac{M+1}{M}, what is ana_n?

Hint

Review Question 3 for using the sequence of partial sums.

Answer

an={2n(n−1)(n−2) if n≥3,−52 if n=2,2 if n=1\displaystyle a_n =\begin{cases} \frac{2}{n(n-1)(n-2)} &\text{ if }n \ge 3,\\ -\frac52 &\text{ if }n=2,\\ 2 &\text{ if }n=1 \end{cases}

Full solution

Since {SM}\{\mathscr S_M\} is the sequence of partial sums of ∑N=1∞SN\displaystyle\sum_{N=1}^\infty S_N, we can find {SN}\{S_N\} from {SM}\{\mathscr S_M\} as in Question 3:

SN=SN−SN−1=N+1N−NN−1=−1N(N−1)if N≥2,S1=S1=2\begin{align*}S_N&=\mathscr S_{N}-\mathscr S_{N-1}=\frac{N+1}{N} - \frac{N}{N-1}=-\frac{1}{N(N-1)} &\text{if }N \geq 2,\\ S_1&=\mathscr S_1 = 2\end{align*}

Similarly, we find {an}\{a_n\} from {SN}\{S_N\}. Do be careful: SNS_N only follows the formula we found above when N≥2N \geq 2. In the next line, we use an expression containing Sn−1S_{n-1}; in order for the subscript to be at least two (so the formula fits), we need n≥3n \geq 3.

an=Sn−Sn−1=−1n(n−1)−−1(n−1)(n−2)=2n(n−1)(n−2)if n≥3,a2=S2−S1=−12(2−1)−2=−52,a1=S1=2\begin{align*}a_n&=S_n-S_{n-1}=-\frac{1}{n(n-1)}-\frac{-1}{(n-1)(n-2)}\\ &=\frac{2}{n(n-1)(n-2)} &\text{if }n \ge 3,\\ a_2&=S_2-S_1=-\frac{1}{2(2-1)} - 2=-\frac52,\\ a_1&=S_1=2\end{align*}

All together,

an={2n(n−1)(n−2) if n≥3,−52 if n=2,2 if n=1\begin{align*}a_n &=\begin{cases} \frac{2}{n(n-1)(n-2)} &\text{ if }n \ge 3,\\ -\frac52 &\text{ if }n=2,\\ 2 &\text{ if }n=1 \end{cases}\end{align*}
Q33Stage 3

Create a bullseye using the following method:

Starting with a red circle of area 1, divide the radius into thirds, creating two rings and a circle. Colour the middle ring blue.

Continue the pattern with the inside circle: divide its radius into thirds, and colour the middle ring blue.

Figure from prob_s3.2, line 2

Figure from prob_s3.2, line 2

Figure from prob_s3.2, line 10

Figure from prob_s3.2, line 10

Continue in this way indefinitely: dividing the radius of the innermost circle into thirds, creating two rings and another circle, and colouring the middle ring blue.

Figure from prob_s3.2, line 2

Figure from prob_s3.2, line 2

What is the area of the red portion?

Hint

What is the ratio of areas between the outermost (red) ring and the next (blue) ring?

Answer

58\dfrac{5}{8}

Full solution

We consider a circle of radius RR, with an “inner ring" from R3\frac{R}{3} to 2R3\frac{2R}{3} and an “outer ring" from 2R3\frac{2R}{3} to RR.

The area of the outer ring is:

πR2−π(2R3)2=59πR2\pi R^2 - \pi \left(\frac{2R}{3}\right)^2=\frac{5}{9}\pi R^2

The area of the inner ring is:

π(2R3)2−π(R3)2=39πR2\pi \left(\frac{2R}{3}\right)^2 - \pi \left(\frac{R}{3}\right)^2=\frac{3}{9}\pi R^2

So, the ratio of the inner ring's area to the outer ring's area is 35\frac{3}{5}.

In our bullseye diagram, if we pair up any red ring with the blue ring just inside it, the blue ring has 35\frac{3}{5} the area of the red ring. So, the blue portion of the bullseye has 35\frac{3}{5} the area of the red portion.

Since the circle has area 1, if we let the red portion have area AA, then

1=A+35A=85A1=A+\frac{3}{5}A=\frac{8}{5}A

So, the red portion has area 58\frac{5}{8}.

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From the UBC Math 101 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.