수열의 극한 전체 문제

Since ${(1+1/n)}^{n}\to e,$ one has ${(1-2/n)}^{n}\approx {(1+k)}^{-2k}\to {e}^{-2}$ as $k\to \infty .$

${2}^{n}+{3}^{n}\le 2\cdot {3}^{n}$ and ${3}^{n}/{4}^{n}\to 0$ as $n\to \infty ,$ so ${a}_{n}\to 0$ as $n\to \infty .$

$\frac{{a}_{n+1}}{{a}_{n}}=n!/(n+1)(n+2)\cdots \,(2n)$ $=\frac{1\cdot 2\cdot 3\cdots n}{(n+1)(n+2)\cdots \,(2n)}<1/{2}^{n}.$ In particular, ${a}_{n+1}/{a}_{n}\le 1/2,$ so ${a}_{n}\to 0$ as $n\to \infty .$

${t}_{k}$ converges to $0.57721\ldots {t}_{k}$ is a sum of rectangles of height $1/k$ over the interval $[k,k+1]$ which lie above the graph of $1/x.$

A sketch of one proof method:

Let any $\epsilon>0$ be given. Since $\{a_n\}$ and $\{b_n\}$ converge, there exists an $N>0$ such that for all $n\geq N$, both $a_n$ and $b_n$ are within $\epsilon/2$ of $L$; we can conclude that they are at most $\epsilon$ apart from each other. Since $a_n\leq c_n \leq b_n$, one can show that $c_n$ is within $\epsilon$ of $L$, showing that $\{c_n\}$ also converges to $L$.

(a) The $n^\text{th}$ partial sum of the odd series is $1+\frac13+\frac15+\cdots+\frac{1}{2n-1}$. The $n^\text{th}$ partial sum of the even series is $\frac12+\frac14 + \frac16 + \cdots +\frac1{2n}$. Each term of the even series is less than the corresponding term of the odd series, giving us our result. (b) The $n^\text{th}$ partial sum of the odd series is $1+\frac13+\frac15+\cdots+\frac1{2n-1}$. The $n^\text{th}$ partial sum of 1 plus the even series is $1+\frac12+\frac14+\cdots + \frac{1}{2(n-1)}$. Each term of the even series is now greater than or equal to the corresponding term of the odd series, with equality only on the first term. This gives us the result. (c) If the odd series converges, the work done in (a) shows the even series converges also. (The sequence of the $n^\text{th}$ partial sum of the even series is bounded and monotonically increasing.) Likewise, (b) shows that if the even series converges, the odd series will, too. Thus if either series converges, the other does.

Similarly, (a) and (b) can be used to show that if either series diverges, the other does, too. (d) If both the even and odd series converge, then their sum would be a convergent series. This would imply that the Harmonic Series, their sum, is convergent. It is not. Hence each series diverges.

(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.

(a) When $a_1=4$, we see $a_2=\dfrac{4+8}{3}=4$, and so on. That is, $a_n=4$ for every $n$. So, $\lim\limits_{n \to \infty} a_n=4$. (b) Cross-multiplying, we see $3x=x+8$, hence $x=4$. (c) In order for our sequence to converge to 4, the terms should be getting infinitely close to 4. So, we find the relationship between $a_{n+1}-4$ and $a_n-4$.

$$\begin{aligned} a_{n+1}&= \frac{a_n+8}{3} \\ a_{n+1}-4&= \frac{a_n+8}{3} -4=\frac{ a_n-4}{3}\\ \end{aligned}$$

So, the distance between our sequence terms and the number 4 is decreasing by a factor of 3 each term. This implies that the terms get infinitely close to 4 as $n$ grows. That is, $\lim\limits_{n \to \infty}a_n=4$.