수열의 수렴과 발산

monotonically increasing

monotonically increasing for $n\geq 3$

never monotonic

monotonically decreasing for $n\geq 3$

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.

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.

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

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.

• Since $-1 \leq \sin n \leq 1$ for all $n$, let $b_n = \frac{n^{-1}}{n^2} = \frac{1}{n^3}$ and $c_n = \frac{n}{n^2}=\frac{1}{n}$. Then $b_n \leq a_n \leq c_n$.
• Both $\lim\limits_{n \to \infty}b_n=0$ and $\lim\limits_{n \to \infty}c_n=0$.

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 $n$.

$$\begin{aligned} d_n&=e^{-1/n}=\frac{1}{e^{1/n}}\\ \lim_{n \to \infty }d_n&=\lim_{n \to \infty }\frac{1}{e^{1/n}} = \frac{1}{e^0}=\frac11=1 \end{aligned}$$

First, we note that both numerator and denominator grow without bound. So, we have to decide whether one outstrips the other, or whether they reach a stable ratio.

• Solution 1: Let's try dividing the numerator and denominator by $2^n$ (the dominant term in the denominator; this is the same idea behind factoring out the leading term in rational expressions). $$b_n = \frac{e^n}{2^n+n^2}\left(\frac{\frac{1}{2^n}}{\frac{1}{2^n}}\right) = \frac{\left(\frac{e}{2}\right)^n}{1+\frac{n^2}{2^n}}$$ Since $e>2$, we see $\dfrac{e}{2}>1$, and so $\lim\limits_{n \to \infty}\left(\dfrac{e}{2}\right)^n=\infty$. Since exponential functions grow much, much faster than polynomial functions, we also see $\lim\limits_{n \to \infty}\frac{n^2}{2^n}=0$. So, $$\lim\limits_{n \to \infty}b_n = \lim\limits_{n \to \infty} \frac{\left(\frac{e}{2}\right)^n}{1+\frac{n^2}{2^n}} = \lim\limits_{n \to \infty} \frac{\left(\frac{e}{2}\right)^n}{1+0} = \infty$$
• Solution 2: Since the numerator and denominator both increase without bound, we apply l'Hopital's rule. Recall $\frac{\mathrm{d}}{\mathrm{d}x}\{2^x\} = 2^x\log 2$.

$$\begin{aligned} \lim_{n \to \infty}b_n&=\lim_{n \to \infty}\underbrace{\dfrac{e^n}{2^n+n^2}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to \infty}{\mathrm{den}\to\infty}}\\ &=\lim_{n \to \infty}\underbrace{\frac{e^n}{2^n\log 2 + 2n}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to \infty}{\mathrm{den}\to\infty}}\\ &=\lim_{n \to \infty} \underbrace{\frac{e^n}{2^n(\log 2)^2 + 2}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to\infty}{\mathrm{den}\to\infty}}\\ &=\lim_{n \to \infty}\frac{e^n}{2^n(\log 2)^3}\\ &=\frac{1}{(\log 2)^3}\lim_{n \to \infty}{\left(\frac{e}{2}\right)^n}\\ &=\infty \end{aligned}$$

Since $e>2$, we see $\dfrac{e}{2}>1$, and so $\lim\limits_{n \to \infty}\left(\dfrac{e}{2}\right)^n=\infty$.

First, we simplify. Remember $n! = n(n-1)(n-2)\cdots(2)(1)$ for any whole number $n$, so $(k+1)!=(k+1)k!$ .

$$\begin{aligned} a_k&=\frac{k!\sin^3k}{(k+1)!}=\frac{k!\sin^3k}{(k+1)k!} = \frac{\sin^3 k}{k+1} \end{aligned}$$

Now, we can use the squeeze theorem.

• $-1 \leq \sin k \leq 1$ for all $k$, so $-1 \leq \sin^3k \leq 1$. Let $b_k = \frac{-1}{k+1}$ and $c_k = \frac{1}{k+1}$. Then $b_k \leq a_k \leq c_k$.
• Both $\lim\limits_{k \to \infty}b_k=0$ and $\lim\limits_{k \to \infty}c_k=0$.

So, by the squeeze theorem, also $\displaystyle\lim_{k\to\infty} a_k=0$.

Note $\lim\limits_{n\to \infty} (-1)^n$ doesn't exist, but $-1 \leq (-1)^n\leq 1$ for all $n$. Let's use the squeeze theorem.

• Let $a_n=-\sin\left(\frac{1}{n}\right)$ and $b_n=\sin\left(\frac{1}{n}\right)$. Then $a_n\leq (-1)^n\sin\left(\frac{1}{n}\right) \leq b_n$.
• Both $\lim\limits_{n \to \infty} -\sin\left(\frac{1}{n}\right)=0$ and $\lim\limits_{n \to \infty} \sin\left(\frac{1}{n}\right)=0$, since $\lim\limits_{n \to \infty} \frac{1}{n}=0$ and $\sin 0=0.$

By the squeeze theorem, the sequence $\left\{(-1)^n\sin\frac{1}{n}\right\}$ converges to $0$.