Differentiability of the sum of the series $\sum_k \sin(kx)/k^2$ How to show the following:
If $ f(x) = \displaystyle\sum_{k=1}^{\infty} \dfrac {\sin(kx)}{k^2} $, then show that $f(x)$ is differentiable on $(0,1)$
I guess it should be related to uniform convergence of partial sums, but how to proceed?
 A: To showing differentiability,  show that $\displaystyle\sum\frac{\cos(kx)}{k}$ converge uniformly on any compact $\subset  (0,1)$.
A: Consider the series of derivatives
$$\sum_{k=1}^{\infty} \frac{\cos kx}{k}.$$
The sequence $1/k$ converges uniformly and montonically to $0$.  The partial sums of the cosine terms are for $x \neq 2m\pi,$ 
$$S_n = \sum_{k=1}^{n} \cos kx= \frac{\sin (\frac{nx}{2})\cos \left(\frac{n+1}{2}x\right)}{\sin(x/2)}$$
and can be shown to be uniformly bounded for all $n$ and $x \in [\delta,1]$ with $0 < \delta <1$. (The closed form for the partial sum on the RHS is derived below).
For $x \in [\delta,1]$, 
$$|S_n| \leq \frac{\left|\sin (\frac{nx}{2})\right|\left|\cos \left(\frac{n+1}{2}x\right)\right|}{\left|\sin(x/2)\right|} \leq \frac{1}{\sin(\delta/2)}.$$
By the Dirichlet test, the series is uniformly convergent on $[\delta,1]$.
Hence, the series
$$\sum_{k=1}^{\infty} \frac{\sin kx}{k^2}$$
can be differentiated term-wise in $[\delta,1].$
Derivation of Partial Sum
Note that
$$\sin\left[\left(k+\frac1{2}\right)x\right]-\sin\left[\left(k-\frac1{2}\right)x\right]=2\cos(kx)\sin(x/2).$$
Hence
$$2\sin(x/2)\sum_{k=1}^{n} \cos kx=\sin\left[\left(n+\frac1{2}\right)x\right]-\sin(x/2),$$
and
$$\sum_{k=1}^{n} \cos kx=\frac{\sin\left[\left(n+\frac1{2}\right)x\right]-\sin(x/2)}{2\sin(x/2)}\\=\frac{\sin\left[\frac{nx}{2}+\left(\frac{n+1}{2}\right)x\right]-\sin\left[\left(\frac{n+1}{2}\right)x-\frac{nx}{2}\right]}{2\sin(x/2)}\\=\frac{\sin (\frac{nx}{2})\cos \left(\frac{n+1}{2}x\right)}{\sin(x/2)}.$$
