What is the value of $\sum e^{-n} \sin^2 n$? Clearly the series $\sum_1^\infty e^{-n} \sin^2 n$ converges. If I put it into Maple, I get an exact value:
$$
-\frac {{\rm e}^1 ( {\rm e}^1 (\cos(1))^2 + (\cos(1))^2 - {\rm e}^1 -1 ) }{-4{\rm e}^2 ( \cos(1))^2+ {\rm e}^3 + 4{\rm e}^1 (\cos(1))^2 + {\rm e}^2 - {\rm e}^1-1}.
$$
How can one derive this by hand?
Does the fact that $\sum_1^\infty e^{-n} \sin^2 n + \sum_1^\infty e^{-n} \cos^2 n$ is computable somehow help? 
 A: $$\sum e^{-n}\left(\frac{e^{in}-e^{-in}}{2i}\right)^2=-\frac14\sum\left(e^{(-1+2i)n}-2e^{-n}+e^{(-1-2i)n}\right).$$
The rest is tedious but straightforward.
A: Hint. Observe that
$$
\sum_{n=0}^{\infty}z^n=\frac1{1-z}, \quad |z|<1. \tag1
$$ Then, put $z:=e^{-1}e^{it}$, with $t \in \mathbb{R}$ in $(1)$ to obtain
$$
\begin{align}
\sum_{n=0}^{\infty}e^{-n}e^{int}=\frac1{1-e^{-1}e^{it}}, \quad t \in \mathbb{R}, \tag2
\end{align}
$$ giving, considering  real parts,
$$
\begin{align}
\sum_{n=0}^{\infty}e^{-n}\cos (nt)&=\Re\frac1{1-e^{-1}e^{it}}\\\\
&=\Re\frac{1-e^{-1}e^{-it}}{(1-e^{-1}e^{it})(1-e^{-1}e^{-it})}\\\\
&=\frac{1-e^{-1}\cos t}{1+e^{-2}-2e^{-1}\cos t} \tag3\\\\
\end{align}
$$ and use $(3)$ with $t=2$ to evaluate your initial series.
A: Note that
$$
\sum_{n=1}^\infty e^{-n} \sin^2 n = 
\frac 12 \sum_{n=1}^\infty e^{-n}(1 - \cos(2n)) = 
\frac 12 \sum_{n=1}^\infty e^{-n} - 
\frac 12 \sum_{n=1}^\infty e^{-n}\cos(2n)
$$
Note also that
$$
e^{-n} \cos(2n) = \Re\left(\left[e^{-1 - 2i} \right]^n\right)
$$
All together, we have
$$
\sum_{n=1}^\infty e^{-n} \sin^2 n = 
\frac 12 \frac{1}{e - 1} - \Re\left[ \frac{1}{e^{1 + 2i} - 1} \right]
$$
