Closed form of $\sum_{n=1}^{\infty} \frac{1}{2^n(1+n^2)}$ How would you recommend me to tackle the series 
$$\sum_{n=1}^{\infty} \frac{1}{2^n(1+n^2)}$$?
Can we possibly express it in terms of known constants? What do you think about it?
 A: You may recall the Lerch transcendent function
$$
\Psi(z,s,a)=\sum_{k=0}^\infty\frac{z^k}{(a+k)^s}
$$
and use
$$
\frac{1}{n^2+1}=\frac{i}{2}\left(\frac{1}{n+i}-\frac{1}{n-i}\right)
$$
to get
$$
\sum_{n=1}^{\infty} \frac{1}{2^n(1+n^2)}=-1-\Im \: \Psi\left(\frac12,1,i\right)
$$
which gives your series in terms of a known special function. 
Maybe someday we will be able to say something deeper...
A: Mathematica yields
Sum[1/(2^n (1 + n^2)), {n, \[Infinity]}]

giving $$\left(\frac{1}{4}-\frac{i}{4}\right) (\, _2F_1(1,1;2-i;-1)+i \, _2F_1(1,1;2+i;-1))\approx0.318057$$
where $_2F_1$ is the confluent 2F1 hypergeometric function, which agrees with numerical summation.
As for how one could obtain this result by hand, perhaps another user could supply that answer?
A: The most compact form I was able to specify is
$$3\,\Re\left({_2F_1}\left(\begin{array}c 1,1 \\ 1+i\end{array}\middle|\,-1\right)\right) - \Im\left({_2F_1}\left(\begin{array}c 1,1 \\ 1+i\end{array}\middle|\,-1\right)\right) - 2\,\Re\left({_2F_1}\left(\begin{array}c 1,1 \\ i\end{array}\middle|\,-1\right)\right).$$
By the way for the numerical value ISC gives back the sum.
