절대수렴과 조건수렴

Note that $(-1)^{2n+1}=(-1)\cdot(-1)^{2n}=-1$. So we can simplify

$$\begin{aligned} \sum_{n=1}^\infty\frac{(-1)^{2n+1}}{1+n}=-\sum_{n=1}^\infty\frac{1}{1+n} \end{aligned}$$

Since $\displaystyle \frac{1}{1+n} \geq \frac{1}{n+n} = \frac{1}{2n}$, $\ \ \displaystyle \sum_{n=1}^\infty\frac{1}{1+n}$ diverges by the comparison test with the divergent harmonic series $\sum\limits_{n=1}^\infty\frac{1}{n}$. The extra overall factor of $-1$ in the original series does not change the conclusion of divergence.

Since

$$\begin{aligned} \lim_{n\to\infty} \frac{1+4^n}{3+2^{2n}} =\lim_{n\to\infty} \frac{1+4^n}{3+4^n} = 1 \end{aligned}$$

the alternating series test cannot be used. Indeed, $\displaystyle \lim_{n\to\infty} (-1)^{n-1} \frac{1+4^n}{3+2^{2n}}$ does not exist (for very large $n$, $(-1)^{n-1} \frac{1+4^n}{3+2^{2n}}$ alternates between a number close to $+1$ and a number close to $-1$) so the divergence test says that the series diverges. (Note that “none of the above” cannot possibly be the correct answer — every series either converges absolutely, converges conditionally, or diverges.)

We first develop some intuition about $\displaystyle\sum_{n=1}^\infty\left|\frac{n^2-\sin n}{n^6+n^2}\right|$, where we take the absolute value of the summands to consider whether the series converges absolutely. For very large $n$, $n^2$ dominates $\sin n$ and $n^6$ dominates $n^2$ so that

$$\begin{aligned} \left|\frac{n^2-\sin n}{n^6+n^2}\right| \approx \frac{n^2}{n^6} =\frac{1}{n^4} \end{aligned}$$

The series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^4}$ converges by the $p$–test with $p=4>1$. We expect the given series to converge too.

To verify that our intuition is correct, we apply the limit comparison test with

$$\begin{aligned} a_n= \frac{n^2-\sin n}{n^6+n^2} \quad\text{and}\quad b_n= \frac{1}{n^4} \end{aligned}$$

which is valid since

$$\lim_{n\rightarrow\infty} \frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\left|\frac{(n^2-\sin n)}{n^6+n^2}\right|\cdot\frac{n^4}{1} =\lim_{n\rightarrow\infty}\frac{|n^6-n^4\sin n|}{n^6+n^2} =\lim_{n\rightarrow\infty}\frac{1-n^{-2}\sin n}{1+n^{-4}} =1$$

exists and is nonzero. Since the series $\sum\limits_{n=1}^\infty b_n$ converges, the series $\sum\limits_{n=1}^\infty\dfrac{|n^2-\sin n|}{n^6+n^2}$ converges too. Therefore, the series $\sum\limits_{n=1}^\infty\dfrac{n^2-\sin n}{n^6+n^2}$ converges absolutely.

You might think that this series converges by the alternating series test. But you would be wrong. The problem is that $\{a_n\}$ does not converge to zero as $n\rightarrow\infty$, so that the series actually diverges by the divergence test. To verify that the $n^{\text{th}}$ term does not converge to zero as $n\rightarrow\infty$ let's write $a_n= \frac{(2n)!}{(n^2+1)(n!)^2}$ (i.e. $a_n$ is the $n^{\text{th}}$ term without the sign) and check to see whether $a_{n+1}$ is bigger than or smaller than $a_n$.

$$\begin{aligned} \frac{a_{n+1}}{a_n} &=\frac{(2n+2)!}{((n+1)^2+1)((n+1)!)^2} \frac{(n^2+1)(n!)^2}{(2n)!} =\frac{(2n+2)(2n+1)}{(n+1)^2}\frac{n^2+1}{(n+1)^2+1} \\ &=\frac{2(2n+1)}{(n+1)}\frac{1+1/n^2}{(1+1/n)^2+1/n^2} =4\frac{1+1/2n}{1+1/n}\frac{1+1/n^2}{(1+1/n)^2+1/n^2} \end{aligned}$$

So

$$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=4$$

and, in particular, for large $n$, $a_{n+1}>a_n$. Thus, for large $n$, $a_n$ increases with $n$ and so cannot converge to $0$. So the series diverges by the divergence test.

The error in our approximation using through term $N$ is at most $\frac{1}{(2(N+1))!}$. We want $\frac{1}{(2(N+1))!}<\frac{1}{1000}$. By checking small values of $N$, we see that $8!=40320>1000$, so if $N=3$, then $\frac{1}{2(N+1)!}=\frac{1}{40320}<\frac{1}{1000}$. So, for our approximation, it suffices to consider the first four terms of our series.

$$\begin{aligned} \cos(1)&\approx \sum_{N=0}^3 \frac{(-1)^n}{(2n)!} = \frac{1}{0!}-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}\\ &=1-\frac12+\frac{1}{24}-\frac{1}{720}\\ &=\frac{720-360+30-1}{720}=\frac{389}{720} \end{aligned}$$

When we use a calculator, we see

$$\begin{aligned} \frac{389}{720}&=0.5402\overline{77} \\ \cos(1)&\approx 0.540302\\ \cos(1) - \frac{389}{720}&\approx 0.000024528\approx \frac{1}{40770} \end{aligned}$$

So, our error is reasonably close to our bound of $\frac{1}{40320}$, and far smaller than $\frac{1}{1000}$.