3.1 Sequences

Not every limit exists.

(a) $-2$ (b) 0 (c) the limit does not exist

(a) The values of the sequence seem to be getting closer and closer to -2, so we guess the limit of this sequence is -2.
(b) Overall, the values of the sequence seem to be getting extremely close to 0, so we approximate the limit of this sequence as 0. It doesn't matter that the sequence changes signs, or that the numbers are sometimes farther from 0, sometimes closer.
(c) This limit does not exist. The sequence is sometimes 0, sometimes -2, and not consistently staying extremely near to either one.

100 isn't all that big when you're contemplating infinity. (Neither is any other number.)

true

True. We consider the end behaviour of the sequences, which does not depend on any finite number of terms at their beginning.

$\displaystyle\lim_{n \to \infty}a_{2n+5}=\lim_{n \to \infty}a_{n}$

(a) $\dfrac{A-B}{C}$ (b) 0 (c) $\dfrac{A}{B}$

(a) We follow the arithmetic of limits, Theorem 3.1.8 in the CLP-2 text: $\dfrac{A-B}{C}$
(b) Since $\lim\limits_{n \to \infty}{c_n}$ is some real number, and $n$ grows without bound, $\lim\limits_{n \to \infty}\dfrac{c_n}{n}=0$.
(c) We note $\displaystyle\lim_{n \to \infty}a_{2n+5}=\lim_{n \to \infty}a_{n}$, so $\displaystyle\dfrac{a_{2n+5}}{b_n} = \frac{A}{B}$.

The sequence might be defined by different functions when $n$ is large than when $n$ is small.

Two possible answers, of many:

There are many possible answers. One is:

where we have a series that looks different before and after its thousandth term. Note every term is smaller than the term preceding it.

Another sequence with the desired properties is:

When $n \leq 1000$, $a_n \geq \frac{1,002,001}{1000}-2>\frac{1,002,000}{1,000}-2=1000$. That is, $a_n>1000$ when $n \le 1000$. As $n$ gets larger, $a_n$ gets smaller, so $a_{n+1}<a_n$ for all $n$. Finally, $\lim\limits_{n \to \infty} a_n = 0-2=-2$.

Recall $(-1)^n$ is positive when $n$ is even, and negative when $n$ is odd.

One possible answer is $a_n=(-1)^{n} = \{-1, 1,-1,1,-1,1,-1,\ldots\}$.
Another is $a_n=n(-1)^{n} = \{-1, 2,-3,4,-5,6,-7,\ldots\}$.

One possible answer is $a_n=(-1)^{n} = \{-1, 1,-1,1,-1,1,-1,\ldots\}$.
Another is $a_n=n(-1)^{n} = \{-1, 2,-3,4,-5,6,-7,\ldots\}$.

Modify your answer from Question 5, but make the terms approach zero.

One sequence of many possible is $\displaystyle a_n = \frac{(-1)^n}{n} = \left\{-1,\ \frac12,\ -\frac13,\ \frac14,\ -\frac15,\ \frac16,\ \ldots \right\}$.

If the terms of a sequence are alternating sign, but the limit of the sequence exists, the limit must be zero. (If it were a positive number, the negative terms would not get very close to it; if it were a negative number, the positive terms would not get very close to it.)

This gives us the idea to modify an answer from Question 5. One possible sequence:

$(-n)^{-n} = \dfrac{(-1)^n}{n^n}$

Some possible answers:

What might cause your answers in (a) and (b) to differ? Carefully read Theorem 3.1.6 in the CLP-2 text about convergent functions and their corresponding sequences.

(a) $a_n=b_n=h(n)=i(n)$, $c_n = j(n)$, $d_n=f(n)$, $e_n=g(n)$
(b) $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} b_n = \lim\limits_{x \to \infty} h(x)=1$, $\lim\limits_{n \to \infty} c_n=\lim\limits_{n \to \infty} e_n=\lim\limits_{x \to \infty} g(x) = \lim\limits_{x \to \infty} j(x) =0$,
$\lim\limits_{n \to \infty} d_n$, $\lim\limits_{x\rightarrow\infty} f(x)$ and $\lim\limits_{x\rightarrow\infty} i(x)$ do not exist.

You can use the fact that $\pi$ is somewhat close to $\dfrac{22}{7}$, or you can use trial and error.

(a) Some possible answers: $a_{22}\approx -0.99996$, $a_{66}\approx -0.99965$, and $a_{110}\approx -0.99902$.

(b) Some possible answers: $a_{11}\approx 0.0044$, $a_{33}\approx -0.0133$, and $a_{55}\approx 0.0221$.
The integers 11, 33, and 55 were found by approximating $\pi$ by $\dfrac{22}{7}$ and finding when an odd multiple of $\dfrac{11}{7}$ (which is the corresponding approximation of $\dfrac{\pi}{2}$) is an integer.

(c) Some possible answers: $a_{44}\approx 0.9998$, $a_{132}\approx 0.9986$ and $a_{220}\approx 0.09961$.
See the solution for how we found them.

(a) We want to find odd multiples of $\pi$ that are close to integers.

(b) If $x = \dfrac{2k+1}{\vphantom{_1}2}\pi$ for some integer $k$ (that is, $x$ is an odd multiple of $\pi/2$), then $\cos x =0$. So, we can either list out the first few terms of $a_n$ until we find three that are very close to $0$, or we can use our approximation $\pi\approx \dfrac{22\vphantom{^1}}{7}$ to choose values of $n$ that are close to $\dfrac{2k+1}{2}\pi$.

(c) One can use the same strategies as we did for parts (a) and (b). But since we already know some $n$'s with $\cos(n)$ close to $-1$, it's easier to use the trig identity

This identity shows that if $\cos(m)$ is close to $-1$, then $\cos(2m)$ is close to $+1$. So let's try $n=2m = 2\times 22=44$, $2\times 66=132$ and $2\times 110=220$.

They do the trick.

Remark: it is possible to turn the ideas of this question into a rigorous proof that $\lim\limits_{n \to \infty}\cos n$ is undefined.

Remark: This question also hints at a property of the set of all numbers $\cos(n)$, with $n$ running over the integers. Mathematicians say that this set is “dense in the interval $[-1,1]$”. This means that if you pick any number $-1\le r\le 1$, there is an integer $n$ such that $\cos(n)$, while not necessarily being exactly $r$, is as close to $r$ as you like. If you feed “$\cos(n)$ dense” into your favourite search engine, you can find out more.

You can compare the leading terms, or factor a high power of $n$ from the numerator and denominator.

(a) $\infty$ (b) $\dfrac{3}{4}$ (c) 0

When determining the end behaviour of rational functions, recall from last semester that we can either cancel out the highest power of $n$ from the numerator and denominator, or skip this step and compare the highest powers of the numerator and denominator.

This isn't a rational expression, but you can treat it in a similar way. Recall $e<3$.

$\infty$

At first glance, we see both the numerator and denominator grow huge as $n$ increases, so we'll need to think a little further to find the limit.

We don't have a rational function, but we can still divide the top and bottom by $n^e$ to get a clearer picture.

Since $e<3$, we see $3-e$ is positive, so $\lim\limits_{n \to \infty}n^{3-e}=\infty$.

The techniques of evaluating limits of rational sequences are again useful here.

0

This isn't a rational sequence, but factoring out $\sqrt{n}$ from the top and bottom will still clear things up.

Use the squeeze theorem.

0

First, let's start with a tempting fallacy.

The denominator grows without bound, so $\lim\limits_{n \to \infty}\dfrac{\cos(n+n^2)}{n}=0$.

It's certainly true that if the limit of the numerator is a real number, and the denominator grows without bound, then the limit of the sequence is zero. However, in our case, the limit of the numerator does not exist. To apply the limit arithmetic rules from the CLP-2 text (Theorem 3.1.8), our limits must actually exist.

A better reasoning looks something like this:

The denominator grows without bound, and the numerator never gets very large, so $\lim\limits_{n \to \infty}\dfrac{\cos(n+n^2)}{n}=0$.

To quantify this reasoning more precisely, we use the squeeze theorem, Theorem 3.1.10 in the CLP-2 text. There are two parts to the squeeze theorem: finding two bounding functions, and making sure these functions have the same limit.

So, by the squeeze theorem, $\lim\limits_{n \to \infty}\dfrac{\cos(n+n^2)}{n}=0$.

$\displaystyle\frac{1}{n}\leq n^{\sin n}\leq n$

0

The denominator of this sequence grows without bound. The numerator is unpredictable: imagine that $n$ is large. When $\sin n$ is close to $-1$, $n^{\sin n}$ puts a power of $n$ “in the denominator," so we can have $n^{\sin n}$ very close to 0. When $\sin n$ is close to 1, $n^{\sin n}$ is close to $n$, which is large.

To control for these variations, we'll use the squeeze theorem.

So, by the squeeze theorem, $\lim\limits_{n \to \infty}\dfrac{n^{\sin n}}{n^2}=0$ as well.

Remark: we also could have used $b_n=0$ for our lower bound, since $a_n \geq 0$ for all $n$.

$e^{-1/n} = \dfrac{1}{e^{1/n}}$; what happens to $\dfrac{1}{n}$ as $n$ grows?

1

Use the squeeze theorem.

0