Closed form for this summation

Question

How do we prove that $$S=\displaystyle\sum_{n=1}^{\infty} e^{-n} \sin n=\dfrac{e\sin 1}{1+e^2-2e\cos 1}\approx0.419$$

We can write the sum as $$S=\Im \sum_{n=1}^{\infty} e^{(i-1)n}$$

I do not know what to do next. Answers or hints appreciated.

Answer 1

$$ \begin{align} S &=\mathrm{Im}\left(\sum_{n=1}^\infty e^{n(-1+i)}\right)\\ &=\mathrm{Im}\left(\frac{e^{-1+i}}{1-e^{-1+i}}\right)\\ &=\mathrm{Im}\left(\frac{e^{-1+i}}{1-e^{-1+i}}\frac{1-e^{-1-i}}{1-e^{-1-i}}\right)\\ &=\mathrm{Im}\left(\frac{e^{-1+i}-e^{-2}}{1-2e^{-1}\cos(1)+e^{-2}}\right)\\ &=\frac{e^{-1}\sin(1)}{1-2e^{-1}\cos(1)+e^{-2}}\\ &=\frac{e\sin(1)}{e^2-2e\cos(1)+1}\\ \end{align} $$

Answer 2

You're almost there. Use the formula for geometric series, and then compute its imaginary part.