Let $$f(x) = \sum_{n=1}^{\infty} \frac{x^n}{2^n} \cos{nx}$$ Proof that $f$ is differentiable on $(-2,2)$
my approach
let $ m := \frac{x}{2} $ so $m<1$ $$ \left| \frac{x^n}{2^n} \cos{nx} \right| \le m^n \rightarrow 0 $$ So by comparison we have point convergence. But I don't know how to deal with diffentiable..
Series of derivatives, $\sum\limits_{n=1}^\infty \frac{n x^{n - 1} \cos nx - n x^n \sin nx}{2^n}$ converges uniformly on any subsegment of $(-2, 2)$ (per comparison again) and series itself converges. Then the series converges to differentiable function (and it's derivative is sum of series of derivatives) - see, for example, theorem 7.17 of Rudin, "Principles of Mathematical Analysis".
To prove the series is uniformly convergent on $(-2 + \alpha, 2 - \alpha)$, we can note that $|n x^{n - 1} \cos nx - nx^n \sin nx| \leqslant 2 \cdot n \cdot (2 - \alpha)^n$, and it is less then $(2 - \frac{\alpha}{2})^n$ for large enough $n$. And the series $\sum_{n=1}^\infty \frac{(2 - \frac{\alpha}{2})^n}{2^n} = \sum\limits_{n=1}^\infty \left(1 - \frac{\alpha}{4}\right)^n$ converges.