Convergence and differentiability of $\ \sum_{n=0}^\infty\cos(\frac{x}{n})-1 $

Question

How to test the convergency of

$$\ \sum_{n=0}^\infty\left(\cos\left(\frac{x}{n}\right)-1\right) $$

I know that after testing whether the sum is convergent (the function is well defined) I can test whether the sum of derivatives of partial sum is uniformally convergent, thus determine whether it is differntiable, however I am stuck on the convergence. I would appreciate any explanation.

Answer 1

$0\leq 1-\cos t \leq \frac {t^{2}} 2$ and $\sum \frac {x^{2}} {n^{2}}$ converge is uniformly convergent on compact sets. Also the differentiated series is uniformly convergent on compact sets (becasue $|\sin (\frac x n )|\leq \frac {|x|} n$). Hence the series converges to a differentiable function.

I am using the following elementary fact:

If $s_n(x) \to f(x)$ uniformly on compact sets and If $s_n'(x) \to g(x)$ uniformly on compact sets then $f$ is differentiable and $f'(x)=g(x)$.