3.4 Absolute and Conditional Convergence

What is conditional convergence?

False. For example, $b_n=\frac{1}{n}$ provides a counterexample.

False. For example if $b_n=\frac{1}{n}$, then $\sum\limits_{n=1}^{\infty}(-1)^{n+1}b_n =\sum\limits_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$ converges by the alternating series test, but $\sum\limits_{n=1}^{\infty} \frac{1}{n}$ diverges by the $p$–test.

Remark: if we had added that $\{b_n\}$ is a sequence of alternating terms, then by Theorem 3.4.2 in the CLP-2 text, the statement would have been true. This is because $\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}b_n$ would either be equal to $\displaystyle\sum_{n=1}^{\infty} |b_n|$ or $-\displaystyle\sum_{n=1}^{\infty} |b_n|$.

If $\sum |a_n|$ converges, then $\sum a_n$ is guaranteed to converge as well.
(That's Theorem 3.4.2 in the CLP-2 text.) So, one of the blank spaces describes an impossible sequence.

Absolute convergence describes the situation where $\sum |a_n|$ converges (see Definition 3.4.1 in the CLP-2 text). By Theorem 3.4.2 in the CLP-2 text, this guarantees that also $\sum a_n$ converges.

Conditional convergence describes the situation where $\sum |a_n|$ diverges but $\sum a_n$ converges (see again Definition 3.4.1 in the CLP-2 text).

If $\sum a_n$ diverges, we just say it diverges. The reason is that if $\sum a_n$ diverges, we automatically know $\sum |a_n|$ diverges as well, so there's no need for a special name.

conditionally convergent

The series $\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{9n+5}$ converges by the alternating series test. On the other hand the series $\sum\limits_{n=1}^{\infty}\big|\frac{(-1)^n}{9n+5}\big| =\sum_{n=1}^{\infty}\frac{1}{9n+5}$ diverges by the limit comparison test with $b_n=\frac{1}{n}$. So the given series is conditionally convergent.

Be careful about the signs.

The series diverges.

Note that $(-1)^{2n+1}=(-1)\cdot(-1)^{2n}=-1$. So we can simplify

Since $\displaystyle \frac{1}{1+n} \geq \frac{1}{n+n} = \frac{1}{2n}$, $\displaystyle \sum_{n=1}^\infty\frac{1}{1+n}$ diverges by the comparison test with the divergent harmonic series $\sum\limits_{n=1}^\infty\frac{1}{n}$. The extra overall factor of $-1$ in the original series does not change the conclusion of divergence.

Does the alternating series test really apply?

It diverges.

Since

the alternating series test cannot be used. Indeed, $\displaystyle \lim_{n\to\infty} (-1)^{n-1} \frac{1+4^n}{3+2^{2n}}$ does not exist (for very large $n$, $(-1)^{n-1} \frac{1+4^n}{3+2^{2n}}$ alternates between a number close to $+1$ and a number close to $-1$) so the divergence test says that the series diverges. (Note that “none of the above” cannot possibly be the correct answer — every series either converges absolutely, converges conditionally, or diverges.)

What does the summand look like when $n$ is very large?

It converges absolutely.

First, we'll develop some intuition. For very large $n$

since $\left|\cos(n)\right| \leq 1$ for all $n$. By the $p$-test, which is in Example 3.3.6 in the CLP-2 text, the series $\displaystyle \sum_{n=5}^\infty \frac{1}{n^p}$ converges for all $p>1$. So we would expect the given series to converge absolutely.

Now, to confirm that our intuition is correct, we'll first try the limit comparison theorem, which is Theorem 3.3.11 in the CLP-2 text, with $a_n=\left| \frac{\sqrt{n}\cos(n)}{n^2-1}\right|$ and $b_n = \frac{1}{n^{3/2}}$.

Unfortunately, this limit doesn't exist, so this attempt to use the limit comparison theorem has failed. Fortunately, having seen that the $\cos n$ caused the failure, it is not hard to adjust our strategy to get a successful proof of absolute convergence. First, in step 1 below, we use the comparison test to eliminate the $\cos n$ and then, in step 2 below, we apply the limit comparison test.

What does the summand look like when $n$ is very large?

It converges absolutely.

We first develop some intuition about $\displaystyle\sum_{n=1}^\infty\left|\frac{n^2-\sin n}{n^6+n^2}\right|$, where we take the absolute value of the summands to consider whether the series converges absolutely. For very large $n$, $n^2$ dominates $\sin n$ and $n^6$ dominates $n^2$ so that

The series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^4}$ converges by the $p$–test with $p=4>1$. We expect the given series to converge too.

To verify that our intuition is correct, we apply the limit comparison test with

which is valid since

exists and is nonzero. Since the series $\sum\limits_{n=1}^\infty b_n$ converges, the series $\sum\limits_{n=1}^\infty\dfrac{|n^2-\sin n|}{n^6+n^2}$ converges too. Therefore, the series $\sum\limits_{n=1}^\infty\dfrac{n^2-\sin n}{n^6+n^2}$ converges absolutely.

This is a trick question. Be sure to verify all of the hypotheses of any convergence test you apply.

It diverges.

You might think that this series converges by the alternating series test. But you would be wrong. The problem is that $\{a_n\}$ does not converge to zero as $n\rightarrow\infty$, so that the series actually diverges by the divergence test. To verify that the $n^{\text{th}}$ term does not converge to zero as $n\rightarrow\infty$ let's write $a_n= \frac{(2n)!}{(n^2+1)(n!)^2}$ (i.e. $a_n$ is the $n^{\text{th}}$ term without the sign) and check to see whether $a_{n+1}$ is bigger than or smaller than $a_n$.

So

and, in particular, for large $n$, $a_{n+1}>a_n$. Thus, for large $n$, $a_n$ increases with $n$ and so cannot converge to $0$. So the series diverges by the divergence test.

Try the substitution $u=\log x$.

It converges absolutely.

This series converges by the alternating series test. We want to know whether it converges absolutely, so we consider the seris$\displaystyle\sum_{n=2}^\infty\left|\frac{(-1)^n}{n(\log n)^{101}}\right|=\sum_{n=2}^\infty\frac{1}{n(\log n)^{101}}$.

We've seen similar function before (e.g. Example 3.3.7 in the CLP-2 text, with $p=101>1$) and it yields nicely to the integral test. Let $f(x) = \frac{1}{x(\log x)^{101}}$. Note $f(x)$ is positive and decreasing for $n \ge 3$. Then by the integral test, the series $\sum_{n=2}^\infty\frac{1}{n(\log n)^{101}}$ converges if and only if the integral $\int_2^\infty \frac{1}{x(\log x)^{101}}\dee{x}$ does. We evaluate the integral using the substitution $u=\log x$, $\dee{u}=\frac{1}{x}\,\dee{x}$.

Since the integral converges, the series $\sum\limits_{n=2}^\infty\frac{1}{n(\log n)^{101}}$ converges, and therefore the series $\sum\limits_{n=2}^\infty\frac{(-1)^n}{n(\log n)^{101}}$ converges absolutely.

Show that it converges absolutely.

See solution.

The sequence has some positive terms and some negative terms, which limits the tests we can use. However, if we consider the series $\displaystyle\sum_{n=1}^\infty \left|\dfrac{\sin n}{n^2}\right|$, we can use the direct comparison test.

For every $n$, $| \sin n| <1$, so $0\le \left|\dfrac{\sin n}{n^2}\right|<\dfrac{1}{n^2}$. Since $\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2}$ converges, then by the direct comparison test, $\displaystyle\sum_{n=1}^\infty \left|\dfrac{\sin n}{n^2}\right|$ converges as well. Then $\displaystyle\sum_{n=1}^\infty \dfrac{\sin n}{n^2}$ converges absolutely– in particular, it converges.

Use a similar method to Queston 10.

See solution.

The terms of this series are sometimes negative (for odd values of $n$ where $\sin n <\frac{1}{2}$) and sometimes positive. But, they are not strictly alternating, so we can't use the alternating series test. Instead, we use a direct comparison test to show the series converges absolutely.

Since $\displaystyle\sum_{n=1}^{\infty} \left(\frac{3}{8}\right)^n$ converges (it's a geometric sum with $|r|<1$), by the direct comparison test, $\displaystyle\sum_{n=1}^{\infty} \left|\left( \frac{\sin x}{4} -\frac18\right)^n\right|$ converges as well.

Then $\displaystyle\sum_{n=1}^{\infty} \left( \frac{\sin x}{4} -\frac18\right)^n$ converges absolutely–and so it converges.

Show it converges absolutely using a direct comparison test.

See solution.

The terms of this series are sometimes negative and sometimes positive. But, they are not strictly alternating, so we can't use the alternating series test. Instead, we use a direct comparison test to show the series converges absolutely.

The series $\displaystyle\sum_{n=1}^\infty \frac{5}{2^{n+1}}$ converges, because it's a geometric series with $r=\frac{1}{2}$. By the direct comparison test, $\displaystyle\sum_{n=1}^\infty \left|\frac{\sin^2 n - \cos^2 n+\tfrac12}{2^n}\right|$ converges as well. Then $\displaystyle\sum_{n=1}^\infty \frac{\sin^2 n - \cos^2 n+\tfrac12}{2^n}$ converges absolutely, so it converges.

For part (a), replace $n$ by $x$ in the absolute value of the summand. Can you integrate the resulting function?

(a) See the solution. (b) $|S - S_5 | \le 24 \times 36 e^{-6^3}$

(a)

(b) In part (a) we saw that $24n^2 e^{-n^3}$ is positive and decreasing. The limit of this sequence equals $0$ (as can be shown with l'H^opital's Rule, just as we did at the end of the third solution of part (a)). Therefore, we can use the alternating series test, so that the error made in approximating the infinite sum $S= \sum\limits_{n=1}^\infty a_n = \sum\limits_{n=1}^\infty (-1)^{n-1} 24n^2 e^{-n^3}$ by the sum of its first $N$ terms, $S_N=\sum\limits_{n=1}^N a_n$, lies between $0$ and the first omitted term, $a_{N+1}$. If we use $5$ terms, the error satisfies

You don't need to add up very many terms for this level of accuracy.

$\cos 1 \approx \frac{389}{720}$; the actual associated error (using a calculator) is about $0.000025$.

The error in our approximation using through term $N$ is at most $\frac{1}{(2(N+1))!}$. We want $\frac{1}{(2(N+1))!}<\frac{1}{1000}$. By checking small values of $N$, we see that $8!=40320>1000$, so if $N=3$, then $\frac{1}{2(N+1)!}=\frac{1}{40320}<\frac{1}{1000}$. So, for our approximation, it suffices to consider the first four terms of our series.

When we use a calculator, we see

So, our error is reasonably close to our bound of $\frac{1}{40320}$, and far smaller than $\frac{1}{1000}$.

Use the direct comparison test to show that the series converges absolutely.

See solution.

The terms of this series are sometimes negative and sometimes positive. But, they are not strictly alternating, so we can't use the alternating series test. Instead, we use a direct comparison test to show the series converges absolutely.

If $n$ is prime, then

If $n$ is not prime, then

For $n$ sufficiently large, $n^2<e^{n/2}$, so for $n$ sufficiently large,

Since $e>1$, then $\sqrt{e}>1$, so the geometric series $\displaystyle\sum \left(\frac{1}{\sqrt{e}}\right)^n$ has $|r|=r=\frac{1}{\sqrt{e}}<1$, so it converges. By the direct comparison test, $\displaystyle\sum_{n=1}^\infty\left|\dfrac{a_n}{e^n}\right|$ converges as well. Then $\displaystyle\sum_{n=1}^\infty\dfrac{a_n}{e^n}$ converges absolutely, so it converges.