How to study convergence of this series?

Question

I have the following series: \begin{equation} \sum_{n=1}^{\infty}(-1)^n\frac{n\cos(x)^n}{n^2+2} \end{equation} I am restudying series and I'm really rusty, and do not know how to begin to study the convergence of this thing. It will depend of course on $x$. Do I have to set $\cos(x)=z$ and study the following power series, \begin{equation} \sum_{n=1}^{\infty}(-1)^n\frac{nz^n}{n^2+2} \end{equation} trying to find the radius of convergence? Any hints?

Answer 1

If this is a calculus class, it is better to observe that $-1 \leq \cos x \leq 1$, giving four easily discerned behaviours.

• $\cos x = -1$: This is the series $\sum_{n \geq 1} \frac{n}{n^2+2}$, which you have other tools for.
• $-1 < \cos x < 0$: Your series is $\sum_{n \geq 1} \frac{n}{n^2+2} |\cos x|^n$ and $$ 0 < \sum_{n \geq 1} \frac{n}{n^2+2} |\cos x|^n \leq \sum_{n \geq 1} |\cos x|^n \text{,} $$ so your series is a positive series bounded by a geometric series.
• $\cos x = 0$: Easy.
• $0 < \cos x$: Now you have an alternating series.

Answer 2

You need to distinguish 3 cases.

If $\cos x\ne \pm 1$, then you have

$$\Big|(-1)^n \frac{n(\cos x)^n}{n^2+2}\Big|\le |\cos x|^n$$

and

$$\sum_{n=1}^{\infty} |\cos x|^n=\frac{1}{1-|\cos x|}-1,$$

which shows that the series $\sum_{n=1}^{\infty} (-1)^n \frac{n(\cos x)^n}{n^2+2}$ converges absolutely in this case.

If $\cos x=-1$, then you get the series $$\sum_{n=1}^{\infty} \frac{n}{n^2+2},$$ which is divergent (basically the harmonic series).

If $\cos x=1$, then you get the series $$\sum_{n=1}^{\infty} (-1)^n\frac{n}{n^2+2},$$ which is not absolutely convergent. However, this series still converges because you can find $C>0$ such that

$$ \Big|\frac{n}{n^2+2}-\frac{n+1}{(n+1)^2+2} \Big| \le \frac{C}{n^2} \quad\forall n\ge 1.$$