수열의 수렴과 발산

bounded, decreasing for $n\ge 2$

${b}_{1}=0,$ ${x}_{1}=2/3,$ ${b}_{2}=1,$ ${x}_{2}=4/3-1=1/3,$ so the pattern repeats, and $1/3=0.010101\ldots .$

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 $0$.