절대수렴과 조건수렴

The series $\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{9n+5}$ converges by the alternating series test. On the other hand the series $\sum\limits_{n=1}^{\infty}\big|\frac{(-1)^n}{9n+5}\big| =\sum_{n=1}^{\infty}\frac{1}{9n+5}$ diverges by the limit comparison test with $b_n=\frac{1}{n}$. So the given series is conditionally convergent.

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 sequence has some positive terms and some negative terms, which limits the tests we can use. However, if we consider the series $\displaystyle\sum_{n=1}^\infty \left|\dfrac{\sin n}{n^2}\right|$, we can use the direct comparison test.

For every $n$, $| \sin n| <1$, so $0\le \left|\dfrac{\sin n}{n^2}\right|<\dfrac{1}{n^2}$. Since $\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2}$ converges, then by the direct comparison test, $\displaystyle\sum_{n=1}^\infty \left|\dfrac{\sin n}{n^2}\right|$ converges as well. Then $\displaystyle\sum_{n=1}^\infty \dfrac{\sin n}{n^2}$ converges absolutely– in particular, it converges.

The terms of this series are sometimes negative and sometimes positive. But, they are not strictly alternating, so we can't use the alternating series test. Instead, we use a direct comparison test to show the series converges absolutely.

$$\begin{aligned} \left|\frac{\sin^2 n - \cos^2 n+\tfrac12}{2^n}\right| \le\frac{1 + 1 +\tfrac12}{2^n} =\frac{5}{2^{n+1}} \end{aligned}$$

The series $\displaystyle\sum_{n=1}^\infty \frac{5}{2^{n+1}}$ converges, because it's a geometric series with $r=\frac{1}{2}$. By the direct comparison test, $\displaystyle\sum_{n=1}^\infty \left|\frac{\sin^2 n - \cos^2 n+\tfrac12}{2^n}\right|$ converges as well. Then $\displaystyle\sum_{n=1}^\infty \frac{\sin^2 n - \cos^2 n+\tfrac12}{2^n}$ converges absolutely, so it converges.