수열의 수렴과 발산

(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.

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

There are many possible answers. One is: $$a_n = \begin{cases} 3000-n & \text{ if }n \leq 1000\\ -2+\frac{1}{n} & \text{ if }n > 1000 \end{cases}$$ 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: $$a_n=\frac{1,002,001}{n}-2$$ 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, $\lim\limits_{n \to \infty} a_n = 0-2=-2$.

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

(a) Since $-1 \leq \sin n \leq 1$ for all $n$, one potential set of upper and lower bound is $$\dfrac{-1}{n}\le \dfrac{\sin n}{n}\le \dfrac{1}{n}$$ Note $\lim\limits_{n \to \infty}\dfrac{-1}{n} = \lim\limits_{n \to \infty}\dfrac{1}{n}$, so these are valid comparison sequences for the squeeze theorem. (b) Since $-1\leq \sin n \leq 1$ and $-5 \leq -5\cos n \leq 5$ for all $n$, we see

$$\begin{aligned} 7 -1 -5&\leq 7+ \sin n - 5 \cos n \leq 7+{red}1+{blue}5\\ 1&\leq 7+ \sin n - 5 \cos n \leq13 \end{aligned}$$

This gives us the idea to try the bounds $$\dfrac{n^2}{13e^n} \le \dfrac{n^2}{e^n(7+\sin n - 5\cos n)} \le \dfrac{n^2}{e^n}$$ We check that $\lim\limits_{n \to \infty}\dfrac{n^2}{13e^n} = \lim\limits_{n \to \infty}\dfrac{n^2}{e^n}$ (they're both 0–you can verify using l'Hopital's rule), so these are indeed reasonable bounds to choose to use with the squeeze theorem. Alternatively, since $0\le \frac{n^2}{13e^n}$, we can also use $$0 \le \dfrac{n^2}{e^n(7+\sin n - 5\cos n)} \le \dfrac{n^2}{e^n}$$ (c) Since $(-n)^{-n} = \dfrac{1}{(-n)^n} = \dfrac{(-1)^n}{n^n}$, we see $$\dfrac{-1}{n^n} \le (-n)^{-n} \le \dfrac{1}{n^n}$$ Since both $\lim\limits_{n \to \infty}\dfrac{-1}{n^n}$ and $\lim\limits_{n \to \infty}\dfrac{1}{n^n}$ are 0, these are reasonable bounds to use with the squeeze theorem.

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

• Solution 1: One way to do that is to remember that $\pi$ is somewhat close to $\dfrac{22}{7}$. Then when we multiply $\pi$ by a multiple of 7, we should get something close to an integer. In particular, $7\pi$, $21\pi$, and $35\pi$ should be reasonably close to $7\left(\dfrac{22}{7}\right)=22$, $21\left(\dfrac{22}{7}\right)=66$, and $35\left(\dfrac{22}{7}\right)=110$, respectively. We check whether they are close enough:

$$\begin{aligned} 7\pi&\approx 21.99 &\qquad 21\pi &\approx 65.97 & \qquad 35\pi &\approx 109.96 \end{aligned}$$

So indeed, $22$, $66$, and $110$ are all within 0.1 of some odd multiple of $\pi$.

Since the cosine of an odd multiple of $\pi$ is $-1$, we expect all of the sequence values to be close to $-1$. Using a calculator:

$$\begin{aligned}a_{22} &= \cos(22) \approx -0.99996,\\ a_{66} &= \cos(66) \approx -0.99965, \\ a_{110} &= \cos(110) \approx -0.99902 \end{aligned}$$

• Solution 2: Alternately, we could have just listed odd multiple of $\pi$ until we found three that are close to integers. $$\begin{array}{c|c} \hline \mathbf{2k+1}&\mathbf{(2k+1)}\boldsymbol{\pi}\\ \hline 1&3.14\\ 3&9.42\\ 5&15.71\\ 7&21.99\\ 9&28.27\\ 11&34.56\\ 13&40.84\\ 15&47.12\\ 17&53.41\\ 19&59.69\\ 21&65.97\\ 23&72.26\\ 25&78.54\\ 27&84.82\\ 29&91.11\\ 31&97.39\\ 33&103.67\\ 35&109.96 \end{array}$$ Some earlier odd multiples of $\pi$ (like $15\pi$ and $29\pi$) get fairly close to integers, but not within 0.1.

(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 $\dfrac{2k+1}{2}\pi$.

• Solution 1: $$\begin{aligned} \dfrac{2k+1}{2}\pi&\approx \frac{(2k+1)\times 22}{2\times 7}=11\times\frac{2k+1}{7} \end{aligned}$$

So, we expect our values to be close to integers when $2k+1$ is a multiple of 7. For example, $2k+1=7$, $2k+1=21$, and $2k+1=35$.

We check: $$\begin{array}{l| l| l} \qquad\mathbf{x}&\mathbf{n} & \mathbf{a_n}\\ \hline ~7\times \dfrac{\pi\vphantom{^A}}{2}\approx 10.99557 & 11&a_{11}\approx0.0044\\[10pt] 21\times \dfrac{\pi}{2} \approx 32.98672&33&a_{33}\approx-0.0133\\[10pt] 35\times \dfrac{\pi}{2} \approx 54.97787&55& a_{55}\approx 0.0221\\ \end{array}$$ These seem like values of $a_n$ that are all pretty close to 0.

• Solution 2: We could have listed the first several values of $a_n$, and looked for some that are close to 0.

$$\begin{array}{c|c} \mathbf{n}&\mathbf{a_n}\\ \hline 1&0.54\\ 2&-0.42\\ 3&-0.99\\ 4&-0.65\\ 5&0.28\\ 6&0.96\\ 7&0.75\\ 8&-0.15\\ 9&-0.91\\ 10&-0.84 \end{array}$$ Oof. Nothing very close yet. Maybe a better way is to list values of $\frac{2k+1}{2}\pi$, and see which ones are close to integers.

$$\begin{array}{c|c} \mathbf{2k+1}&\mathbf{\frac{2k+1}{2}}\boldsymbol{\pi}\\ \hline 1&1.57\\ 3&4.71\\ 5&7.85\\ 7&10.996\\ 9&14.14\\ 11&17.28\\ 13&20.42\\ 15&23.56\\ 17&26.70\\ 19&29.85\\ 21&32.99\\ 23&36.13\\ 25&39.27\\ 27&42.41\\ 29&45.55\\ 31&48.69\\ 33&51.84\\ 35&54.98 \end{array}$$ We find roughly the same candidates we did in Solution 1, depending on what we're ready to accept as “close".

(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

$$\begin{aligned} \cos(2m) = 2\cos^2(m) -1 \end{aligned}$$

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 $2\times 110=220$. $$\begin{array}{c|c} \mathbf{n}&\mathbf{a_n}\\ \hline 44&0.9998\\ 132&0.9986\\ 220&0.9961 \end{array}$$ 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.

• Let, for each integer $k\ge 1$, $n_k$ be the integer that is closest to $2k\pi$. Then $2k\pi-\frac{1}{2}\le n_k \le 2k\pi+\frac{1}{2}$ so that $\cos(n_k)\ge\cos\frac{1}{2}\ge 0.8$. Consequently, if $\lim\limits_{n \to \infty}\cos n=c$ exists, we must have $c\ge 0.8$.
• Let, for each integer $k\ge 1$, $n'_k$ be the integer that is closest to $(2k+1)\pi$. Then $(2k+1)\pi-\frac{1}{2}\le n'_k \le (2k+1)\pi+\frac{1}{2}$ so that $\cos(n'_k)\le-\cos\frac{1}{2}\le -0.8$. Consequently, if $\lim\limits_{n \to \infty}\cos n=c$ exists, we must have $c\le -0.8$.
• It is impossible to have both $c\ge 0.8$ and $c\le -0.8$, so $\lim\limits_{n \to \infty}\cos n$ does not exist.

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.

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.

(a) Since the numerator has a higher degree than the denominator, this sequence will diverge to positive or negative infinity; since its terms are positive for large $n$, its limit is (positive) infinity. (You can imagine that the numerator is growing much, much faster than the denominator, leading the terms to have a very, very large absolute value.)

Calculating the longer way:

$$\begin{aligned} a_n &= \dfrac{3n^2-2n+5}{4n+3}\left(\dfrac{\frac1n}{\frac1n}\right)=\dfrac{3n-2+\frac5n}{4+\frac3n} \\\lim_{n \to \infty}a_n &= \lim_{n \to \infty}\dfrac{3n-2+\frac5n}{4+\frac3n} = \lim_{n \to \infty}\dfrac{3n-2+0}{4+0} = \infty \end{aligned}$$ (b) Since the numerator has the same degree as the denominator, as $n$ goes to infinity, this sequence will converge to the ratio of their leading coefficients: $\dfrac{3}{4}$. (You can imagine that the numerator is growing at roughly the same rate as the denominator, so the terms settle into an almost-constant ratio.)

Calculating the longer way:

$$\begin{aligned} b_n &= \dfrac{3n^2-2n+5}{4n^2+3}\left(\dfrac{\frac{1}{n^2}}{\frac{1}{n^2}}\right)=\dfrac{3-\frac2n+\frac{5}{n^2}}{4+\frac{3}{n^2}} \\\lim_{n \to \infty}b_n &= \lim_{n \to \infty}\dfrac{3-\frac2n+\frac{5}{n^2}}{4+\frac{3}{n^2}} = \frac{3-0+0}{4+0}=\frac{3}{4} \end{aligned}$$ (c) Since the numerator has a lower degree than the denominator, this sequence will converge to 0 as $n$ goes to infinity. (You can imagine that the denominator is growing much, much faster than the numerator, leading the terms to be very, very small.)

Calculating the longer way:

$$\begin{aligned} c_n &= \dfrac{3n^2-2n+5}{4n^3+3}\left(\dfrac{\frac{1}{n^3}}{\frac{1}{n^3}}\right)=\dfrac{\frac{3}{n}-\frac{2}{n^2}+\frac{5}{n^3}}{4+\frac{3}{n^3}} \\\lim_{n \to \infty}c_n &= \lim_{n \to \infty}\dfrac{\frac{3}{n}-\frac{2}{n^2}+\frac{5}{n^3}}{4+\frac{3}{n^3}} = \dfrac{0-0+0}{4+0}=0 \end{aligned}$$

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.

$$\begin{aligned} a_n&=\dfrac{4n^3-21}{n^e+\frac{1}{n}}\left(\frac{\frac{1}{n^e}}{\frac{1}{n^e}}\right) = \dfrac{4n^{3-e}-\frac{21}{n^e}}{1+\frac{1}{n^{e+1}}} \\ \text{Since $e<3$, we see $3-e$ is positive, so $\lim\limits_{n \to \infty}n^{3-e}=\infty$.} \\ \lim_{n \to \infty}a_n&=\lim_{n \to \infty} \dfrac{4n^{3-e}-\frac{21}{n^e}}{1+\frac{1}{n^{e+1}}} =\lim_{n \to \infty} \dfrac{4n^{3-e}-0}{1+0} = \infty \end{aligned}$$

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

$$\begin{aligned} b_n &= \dfrac{\sqrt[4]{n}+1}{\sqrt{9n+3}}\left(\dfrac{\frac{1}{\sqrt n}}{\frac{1}{\sqrt n}}\right)= \dfrac{\frac{1}{\sqrt[4]{n}}+\frac{1}{\sqrt{n}}}{\sqrt{9+\frac{3}{n}}}\\ \lim_{n \to \infty}b_n &= \lim_{n \to \infty} \dfrac{\frac{1}{\sqrt[4]{n}}+\frac{1}{\sqrt{n}}}{\sqrt{9+\frac{3}{n}}} = \dfrac{0+0}{\sqrt{9+0}}=0 \end{aligned}$$