About $\sum_{k=1}^\infty \frac{b_k}{k}$, where $b_k$ are Fourier coefficients This is my first post here.  
I have some troubles with this property of the Fourier coefficients. Indeed, let $f(x)$ be a continuous real function, with compact support $[a,b] \subset (0,2\pi)$, and
$$
b_k := \frac{1}{\pi}\int_0^{2\pi}f(x)\sin kx \; \mathrm{d}x
$$ 
My question is: does the series $$ \sum_{k=1}^\infty \frac{b_k}{k}$$ converge? If the answer is yes, is its sum equal to $\int_0^{2\pi}\frac{\pi-x}{2}f(x) \; \mathrm{d}x$?
How can I prove/disprove this? What do you suggest? Thanks a lot.
 A: For arbitrary continuous function $\varphi\in C([0,2\pi])$ denote
$$
a_0(\varphi)=\frac{1}{\pi}\int\limits_{0}^{2\pi}\varphi(x)dx
$$
$$
a_k(\varphi)=\int\limits_{0}^{2\pi}\varphi(x)\cos(kx)dx\qquad b_k(\varphi)=\int\limits_{0}^{2\pi}\varphi(x)\sin(kx)dx
$$
Consider function
$$
F(x)=\int\limits_{0}^x f(t)dt-\frac{a_0(f)x}{2}
$$
Since $f\in C([0,2\pi])$  then $F\in C^1([0,2\pi])$. Note that 
$$
F(x+2\pi)-F(x)=\int\limits_{0}^{2\pi}f(t)dt-a_0(f)\pi=0.
$$
So $F$ is $2\pi$ periodic. Since $F$ is $2\pi$ periodic continuously differentiable function its Fourier series uniformly converges to $F$. Thus we have for all $x\in[0,2\pi]$ the following equality
$$
F(x)=\frac{a_0(F)}{2}+\sum\limits_{k=1}^\infty\left(a_k(F)\cos(kx)+b_k(F)\sin(kx)\right)
$$
Integration by parts gives us
$$
a_0(F)=\frac{1}{\pi}\int\limits_{0}^{2\pi}F(x)dx=
\frac{1}{\pi}\left(xF(x)|_0^{2\pi}-\int\limits_{0}^{2\pi}x\left(f(x)-\frac{a_0(f)}{2}\right)dx\right)=
\frac{1}{\pi}\left(-\int\limits_{0}^{2\pi}xf(x)dx+\int\limits_{0}^{2\pi}\pi f(x)dx\right)=
\int\limits_{0}^{2\pi}\frac{\pi-x}{\pi}f(x)dx
$$
$$
a_k(F)=\int\limits_{0}^{2\pi}F(x)\cos(kx)dx=
\left(F(x)\frac{\sin(kx)}{k}\right)_0^{2\pi}-
\int\limits_{0}^{2\pi}\left(f(x)-\frac{a_0(f)}{2}\right)\frac{\sin(kx)}{k}dx=
$$
$$
-\frac{1}{k}\int\limits_{0}^{2\pi}f(x)\sin(kx)dx=-\frac{1}{k}b_k(f)
$$
$$
b_k(F)=\int\limits_{0}^{2\pi}F(x)\sin(kx)dx=
\left(F(x)\frac{-\cos(kx)}{k}\right)_0^{2\pi}-
\int\limits_{0}^{2\pi}\left(f(x)-\frac{a_0(f)}{2}\right)\frac{-\cos(kx)}{k}dx=
$$
$$
\frac{1}{k}\int\limits_{0}^{2\pi}f(x)\cos(kx)dx=\frac{1}{k}a_k(f)
$$
Hence
$$
F(x)=\frac{a_0(F)}{2}+
\sum\limits_{k=1}^\infty\left(-\frac{b_k(f)}{k}\cos(kx)+\frac{a_k(f)}{k}\sin(kx)\right)
$$
Substitute $x=0$, then
$$
0=\frac{a_0(F)}{2}-\sum\limits_{k=1}^\infty\frac{b_k(f)}{k}
$$
i.e.
$$
\sum\limits_{k=1}^\infty\frac{b_k(f)}{k}=
\frac{a_0(F)}{2}=
\int\limits_{0}^{2\pi}\frac{\pi-x}{2\pi}f(x)dx
$$
A: We can weaken the conditions as "$f\in R[0,2\pi]$".In fact ,notice that
$$\frac{\pi-x}2\sim \sum_{n=1}^\infty\frac{\sin nx}n$$
by Generalized Parseval theorem，we immediately get
$$\frac1\pi\int_{0}^{2\pi} \frac{\pi-x}2f(x)dx=\sum_{n=1}^\infty\frac{b_n}n$$
