Does a closed form exist for this summation? How do I calculate $$\sum_{k=1}^\infty \frac{k\sin(kx)}{1+k^2}$$ for $0<x<2\pi$? Wolfram alpha can't calculate it, but the sum surely converges.
 A: We can write the sum as a linear combination of the LerchPhi function 

$$ \Phi(z,s,a) = \sum_{k=0}^{\infty} \frac{z^k}{(k+a)^s}. $$

First, we write the sum as (using partial fraction)
$$ S = \sum_{k=0}^{\infty}\frac{e^{ikx}}{4i(k-i)} -\sum_{k=0}^{\infty} \frac{e^{-ikx}}{4i(k-i)}+ \sum_{k=0}^{\infty}\frac{e^{ikx}}{4i(k+i)}-\sum_{k=0}^{\infty}\frac{e^{-ikx}}{4i(k+i)} $$

$$\implies S =\frac{1}{4i}( \Phi(e^{ix},1,-i)- \Phi(e^{-ix},1,-i)+  \Phi(e^{ix},1,i)- \Phi(e^{-ix},1,i) ).$$

Note: this is a general technique which can handle more general series than the one in considdration.
A: Using known Fourier series expansion $$\sinh(at)=\frac{2\sinh \pi a}{\pi}\sum_{n=1}^\infty\frac{(-1)^{n+1}n\sin nt}{n^2+a^2},\quad t\in(-\pi,\pi)$$
(which can be easily proven simply by calculating the Fourier coefficients for this function) and noting that $(-1)^{n+1}\sin nt=\sin((\pi-t)n)$ we get the series in question taking $a=1$ and $t=\pi-x$: 
$$\sum_{n=1}^\infty\frac{n\sin nx}{n^2+1}=\frac{\pi\sinh(\pi-x)}{2\sinh \pi},\quad x\in(0,2\pi).$$
A: You may use Mathematica.  I tried Mathematica 7.0 and it gives:

Where $_2F_1$ is the hypergeometric function:

