Evaluate the sum $\forall\:\:\alpha,\beta\in\mathbb{R}$ $$S=\sum_{n=1}^{\infty}\frac{\alpha^{n+1}-1}{n(n+1)}\sin\left(\frac{n\pi}{\beta}\right)$$
I rewrote this as $$S=\sum_{n=1}^{\infty}\left(\left(\frac{\alpha^{n+1}-1}{n}-\frac{\alpha^{n+1}-1}{n+1}\right)\sin\left(\frac{n\pi}{\beta}\right)\right)$$ But this is not helping me. I also expanded the sin by its Taylor Series, but it got even more complicated. I then tried wolframaplha, but it does not recognize it as a problem. It doesn't read it as a problem. How should I proceed from here$?$ Any help is greatly appreciated.
Update
With the helpful hints I got, I managed to proceed as follows
$$\frac{\partial^2S}{\partial\alpha^2}=\mathfrak{J}\left(\frac{1}{\alpha}\sum_{n=1}^{\infty}(\alpha\cdot e^{\frac{i\pi}{\beta}})^n\right)$$
Now this is just a normal Geometric Progression. But after this what should I do$?$ How to separate the imaginary part$?$ And, I don't think I'm equipped enough to integrate the resultant expression. Any help is greatly appreciated.
Best Answer
Easier: put $z=e^{\pm i\pi/\beta}$ and $z=\alpha e^{\pm i\pi/\beta}$ into $$\sum_{n=1}^\infty\frac{z^{n+1}}{n(n+1)}=z+(1-z)\log(1-z)\qquad(|z|\leqslant 1)$$