Show that $\sum_{n=1}^{\infty} {(-1)^n \sin(\frac{x}{n})}$ converges for $a > 0$ in $[-a,a]$ Given:
$$\sum_{n=1}^{\infty} {(-1)^n \sin(\frac{x}{n})}$$
We need to show that it converges in $[-a,a]$ $\forall a >0$.
Now what I figured out:

This is a function series summation. we need first to know what $C_n$ actually is and then do the Radius test.

But how do I exactly get it in the general form ?
 A: What you should realise is that $\sum\limits_{n=1}^{\infty}(-1)^{n}\sin\left( \frac{x}{n} \right)$ converges iff $\sum\limits_{n=N+1}^{\infty}(-1)^{n}\sin\left( \frac{x}{n} \right)$ converges
where $N\in\mathbb{N}$ is such that $N \leq \vert a \vert$ and $N+1 > \vert a \vert$ as changing a finite amount of summands has no effect on the convergence. Now you should carefully argue that you may apply the Leibniz criterion. 
A: for $n \ge |a|$,  $\sin \left( \frac{x}{n} \right)$ has same sign as $x$, so we can use Jordan's inequality and then use Leibniz criterion to argue that the series converges.
A: Note that we can write
$$\begin{align}
\sum_{n=1}^N (-1)^n\sin\left(\frac xn\right)&=\sum_{n=1}^N (-1)^n\left(\frac xn\right)+\sum_{n=1}^N (-1)^n\left(\sin\left(\frac xn\right)-\frac xn\right)\\\\
&=x\sum_{n=1}^N \frac{(-1)^n}n+\sum_{n=1}^N (-1)^n\left(\sin\left(\frac xn\right)-\frac xn\right)\tag1
\end{align}$$
The first summation on the right-hand side of $(1)$ is the alternating harmonic sum, which converges as $N\to\infty$.
For the second summation on the right-hand side of $(1)$ we have the estimate
$$\begin{align}
\left|\sum_{n=1}^N (-1)^n\left(\sin\left(\frac xn\right)-\frac xn\right)\right|&\le \sum_{n=1}^N \left|\sin\left(\frac xn\right)-\frac xn\right|\\\\
&\le \frac{|x|^3}6\sum_{n=1}^N\frac1{n^3}
\end{align}$$
Inasmuch as the series $\sum_{n=1}^\infty\frac1{n^3}$ converges, $\sum_{n=1}^\infty (-1)^n\left(\sin\left(\frac xn\right)-\frac xn\right)$ is absolutely convergent.
Putting it all together shows that the series of interest converges for all $x$. 
