Proving series converges using Fourier series I need to prove that for every $0 \le x \le \pi$ $$\sum^{\infty}_{n=-\infty}\frac{e^{2inx}}{1-4n^2}=\frac{\pi}{2}\sin{x}$$ using Fourier series.
Let $f(x)$ be a function such that its Fourier series is
$$f(x)\sim\sum^{\infty}_{n=-\infty}\frac{e^{2inx}}{1-4n^2}=\sum^{\infty}_{n=-\infty}\frac{1}{1-4n^2}e^{2inx}=\sum^{\infty}_{n=-\infty}\frac{1}{1-4n^2}e^{in\pi x/(\pi/2)}$$
So I know $L=\pi/2$, $b_n=-2Im(\frac{1}{1-4n^2})=0$, and $a_n=2Re(\frac{1}{1-4n^2})=\frac{2}{1-4n^2}$ ($a_n$ and $b_n$ are the Fourier coefficients) hence the the series is of the form
$$f(x)\sim \sum^{\infty}_{n=0}\frac{2}{1-4n^2}\cos{2nx}$$
I assume that in order to prove the required convergence I need to find $f(x)$ such that its Fourier series is as described above, but this is a series of an even function(it has only cosine) and $\frac{\pi}{2}\sin{x}$ is clearly an odd function, thus I don't understand how can I find such $f(x)$

For $0\le x\le\pi$,
\begin{align}
\sin(x)=|\sin(x)|&=\frac2{\pi}-\frac4{\pi}\sum_{n=1}^\infty\frac{\cos(2nx)}{4n^2-1}\\
&=\frac2{\pi}-\frac2{\pi}\sum_{n=1}^\infty\frac{\cos(2nx)+\cos(-2nx)+i\sin(2nx)+i\sin(-2nx)}{4n^2-1}\\
&=\frac2{\pi}-\frac2{\pi}\left(1+\sum_{n=-\infty}^\infty\frac{\cos(2nx)+i\sin(2nx)}{4n^2-1}\right)\\
&=\frac2{\pi}\sum_{n=-\infty}^\infty\frac{\cos(2nx)+i\sin(2nx)}{1-4n^2}\\
\end{align}
