For which $x\in \mathbb{R},$ the series $\sum_{n=1}^\infty\frac{\cos^{2n}x} n$ is differentiable? 
Let $\displaystyle\sum_{n=1}^\infty \frac{\cos^{2n}x}{n}$. For which $x\in
\mathbb{R},$ is the series differentiable?

My attempt:
I know the series converges pointwise for every $x\ne \pi\cdot k$, but don't know how to continue from here.
 A: Let $$f_n(x)=\frac{\cos^{2n}x}{n},\space n=1,2,\dots$$  Then $$f'_n(x)=-2\cos^{2n-1}x\sin x,$$ so that $\sum{f_n'(x)}$ converges uniformly on $(-\pi,\pi)$ by the Weierstrass M-test. By the theorem on uniform convergence of the derivatives, applied to the sequence of partial sums, the original series is differentiable on $(-\pi,\pi)$, and by periodicity, for every $x\neq k\pi$.
A: In this case, you don't really need to invoke any theorems, you can just sum the series and work with the function and its derivative directly:  Since $\sum_{n=1}^\infty{u^n\over n}=-\ln(1-u)$ for $|u|\lt1$ and $\cos^2x\lt1$ if $x\not\in\pi\mathbb{Z}$, we have
$$f(x)=\sum_{n=1}^\infty{\cos^{2n}x\over n}=\sum_{n=1}^\infty{(\cos^2x)^n\over n}=-\ln(1-\cos^2x)=-2\ln|\sin x|$$
for $x\not\in\pi\mathbb{Z}$ as well.  It's also easy to check that
$$\sum_{n=1}^\infty\left(\cos^{2n}x\over n\right)'
=-2\sin x\sum_{n=1}^\infty\cos^{2n-1}x
=-2\sin x\cos x\sum_{n=0}^\infty(\cos^2x)^n\\
=-2\sin x\cos x\cdot{1\over1-\cos^2x}
=-2{\cos x\over\sin x}=f'(x)$$
(again for $x\not\in\pi\mathbb{Z}$). So the series is (termwise) differentiable for all $x\not\in\pi\mathbb{Z}$.
