Beginner Fourier Problem and absolute convergence presumably as preparation for Fourier Analysis, I was given the following exercises:

*

*Let $\sum_{n=1}^{\infty} a_{n}$ be a series which is absolute convergent and for $x\in\mathbb{R}$
$$ f(x)=\sum_{n=1}^{\infty} a_{n}\sin(nx)$$
Prove the equation
$$a_{n}=\frac{1}{\pi}\int_{0}^{2\pi} f(x)\sin(nx)\text{d}x\qquad\forall n\in\mathbb{N}$$


*Consider a sequence $ (a_{n})_{n\in\mathbb{N}}$ so that the series
$$\sum_{n=1}^{\infty} n^{2k}a_{n}\qquad \forall n\in\mathbb{N}$$
is absolute convergent. Proof that for $x\in\mathbb{R}$ the function
$$ f(x)=\sum_{n=1}^{\infty} a_{n}\sin(nx)$$
can be differentiated $2k$-times and it holds
$$ f^{(k)}(x)=\sum_{n=1}^{\infty} b_{n}\sin(nx) $$
Determine the coefficient $b_{n}$.
I've spent quite a long time, but unfortunately I don't succeeded.
This is an outline what I've tried so far:

*

*As it holds
$$\left| a_{n}\sin(nx)\right| \leq \left| a_{n}\right|$$
and $\sum_{n=1}^{\infty}\left| a_{n}\right|$ is (absolute) convergent, Weierstrass M-Test gives us that $f(x)$ converge uniformly for $x\in\mathbb{R}$. Therefore
$$  \frac{1}{\pi}\int_{0}^{2\pi} \left(\sum_{n=1}^{\infty} a_{n}\sin(nx)\right) \sin(nx)\text{d}x \\ = \frac{1}{\pi} \sum_{n=1}^{\infty} a_{n} \int_{0}^{2\pi}  \sin^{2}(nx)\text{d}x \\ = \frac{1}{\pi} \sum_{n=1}^{\infty} a_{n} \left[ - \frac{\cos(nx)\sin(nx)}{2n} + \frac{x}{2}\right]^{2\pi}_{0} \\ =\frac{1}{\pi} \sum_{n=1}^{\infty} a_{n} \left[ - \frac{\sin(2nx)}{4n} + \frac{x}{2}\right]^{2\pi}_{0} \\ =\frac{1}{\pi} \sum_{n=1}^{\infty} a_{n} \cdot \left( - \frac{\sin(4\pi n)-4\pi n}{4n}\right) \\ =\frac{1}{\pi} \sum_{n=1}^{\infty} a_{n} \cdot \pi \\ = \sum_{n=1}^{\infty} a_{n} \overset{?}{=}a_{n}$$


*I found no successful method, maybe Taylor's theorem for $\sin$ might be helpful, as it somehow matches to the setup of the exercise.
$$f(x)=\sum_{n=1}^{\infty} a_{n}\sum_{i=0}^{\infty}(-1)^{i}\frac{(nx)^{2i+1}}{(2i+1)!}$$
Thanks in advance for your help.
 A: *

*The series $\sum_{n=1}^\infty a_n$ is absolutely convergent $\implies$ (by Weierstarss M-test) $f(x)=\sum_{n=1}^\infty a_n \sin(nx)$ converges uniformly. Applying the uniform convergence we get that $f$ is absolutely integrable on $[0,2\pi]$ : $$\int_{0}^{2\pi}|f(t)|dt=\int_{0}^{2\pi}|\sum_{n=1}^\infty a_n \sin(nt)|dt \le\sum_{n=1}^\infty|a_n|\int_{0}^{2\pi}|sin(nt)|dt=2\sum_{n=1}^\infty|a_n|<\infty$$
So we can compute it's Fourier coefficients.
Now rewrite, $f(x)=\sum_{n=1}^\infty a_n \sin(nx)=\sum_{n=1}^\infty\frac{a_n}{2i}(e^{inx}-e^{-inx})$
$\hat{f}(n)=\frac{1}{2\pi}\int_{0}^{2\pi} f(x) e^{-inx}dx=\frac{1}{2\pi}\int_{0}^{2\pi}\sum_{m=1}^\infty\frac{a_m}{2i}(e^{imx}-e^{-imx})(e^{-inx})dx=$$\frac{1}{2\pi}\sum_{m=1}^\infty\frac{a_m}{2i}\int_{0}^{2\pi}(e^{imx}-e^{-imx})(e^{-inx})dx$
$$\implies \hat{f}(n)=\frac{a_n}{2i} \frac{1}{2\pi}\int_{0}^{2\pi}dx=\frac{a_n}{2i}$$
Similarly, $$\hat{f}(-n)=-\frac{a_n}{2i}$$
So $a_n=\frac{a_n}{2}-(\frac{-a_n}{2})=i(\hat{f}(n)-\hat{f}(-n))=\frac{1}{\pi}\int_{0}^{2\pi} f(x)\sin(nx)dx$


*$$\forall 0\le j \le2k,\sum_{n=1}^\infty |n^ja_n| \le \sum_{n=1}^\infty n^{2k}|a_n|<\infty \dots (*)$$ .
Hence again Weierstrass M-test tells you that the series $\sum_{n=1}^\infty a_n\sin(nx)$ converges uniformly and also the series made of the derivatives $\sum_{n=1}^\infty na_n\cos(nx)$ converges uniformly. Thus the function defined as the sum $f(x)=\sum_{n=1}^\infty a_n\sin(nx)$ is diffrentiable and moreover, $f^{/}(x)=\sum_{n=1}^\infty na_n\cos(nx)$.

Keeping $(*)$ in mind keep on applying Weierstarss M-test and diffrentiability and computation of derivatives follow. It follows that  $$  f^{(2k)}(x)= \begin{cases}
       \sum_{n=1}^\infty n^{2k} sin(nx) & \text{ if } 2k \equiv 0 \text{ mod 4} \\
            \sum_{n=1}^\infty (-n^{2k}) sin(nx) & \text{ if } 2k \equiv 2 \text{ mod 4} \\
%-\sum_{n=1}^\infty n^{2k} cos(nx) & \text{ if } 2k \equiv 3 \text{ mod 4} \\%
   \end{cases}$$
