Closed form of $\sum_{n=1}^{\infty} \frac{1}{2^n(1+n^2)}$

Question

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?

Answer 1

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?

Answer 2

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...

Answer 3

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.