수열의 극한 전체 문제

(a) With partial fractions, $a_n = \frac32\left(\frac1n-\frac1{n+2}\right)$. Thus $S_n = \frac32\left(\frac32-\frac1{n+1}-\frac1{n+2}\right)$. (b) Converges to 9/4

(a) With partial fractions, $a_n = \frac12\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$. Then $S_n = \frac12\left(1-\frac1{2n+1}\right) = \frac{n}{2n+1}$. (b) Converges to $1/2$.

(a) $S_n = \ln \big(1/(n+1)\big)$ (b) Diverges (to $-\infty$).

(a) $S_n = 1-\frac{1}{(n+1)^2}$ (b) Converges to 1.

(a) $a_n = \frac1{n(n+3)}$; using partial fractions, the resulting telescoping sum reduces to $S_n = \frac13\left(1+\frac12+\frac13-\frac1{n+1}-\frac1{n+2}-\frac1{n+3}\right)$ (b) Converges to $11/18$.

(a) $a_n = 1/2^n+1/3^n$ for $n\geq 0$. Thus $S_n = \frac{1-1/2^2}{1/2}+\frac{1-1/3^n}{2/3}$. (b) Converges to $2+3/2 = 7/2$.

(a) With partial fractions, $a_n = \frac12\left(\frac1{n-1}-\frac1{n+1}\right)$. Thus $S_n = \frac12\left(3/2-\frac1n-\frac{1}{n+1}\right)$. (b) Converges to 3/4.

(a) $S_n=\frac{1-(\sin 1)^{n+1}}{1-\sin 1}$ (b) Converges to $\frac{1}{1-\sin 1}$..

(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) 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".