Computing the infinite sum, $\sum_{n=0}^\infty \frac {5^n}{25^n + 1} $ So I'm trying to compute $$ \displaystyle\sum_{n=0}^\infty \dfrac {5^n}{25^n + 1}. $$ The closed form is not very nice and I don't see any immediate telescoping. Any ideas? 
 A: Since:
$$\frac{5^n}{25^n+1}=\frac{1}{5^n}-\frac{1}{125^n}+\frac{1}{3125^n}-\ldots $$
we have:
$$\sum_{n=0}^{+\infty}\frac{5^n}{25^n+1}=\frac{1}{2}+\left(\frac{1}{4}-\frac{1}{124}+\frac{1}{3124}+\ldots\right)=\frac{1}{2}+\sum_{k=0}^{+\infty}\frac{(-1)^k}{5^{2k+1}-1}.$$

Despite being easy to compute through Euler's acceleration technique, such a series does not have a nice closed expression. Otherwise, also the reciprocal Lucas constant would have one.
A: hint: 
$$\displaystyle\sum_{n=0}^\infty \dfrac {5^n}{25^n + 1} = \displaystyle\sum_{n=0}^\infty \dfrac {1}{5^n + 5^{-n}}=\displaystyle\sum_{n=0}^\infty \dfrac {1}{e^{\ln(5)n} + e^{-\ln(5)n}}=\displaystyle\sum_{n=0}^\infty \dfrac {1}{2\cosh(\ln(5)n)} $$
if you could find a formula for $$\displaystyle\sum_{n=0}^\infty \dfrac {1}{\cosh(xn)} $$
you have soved the problem. and maybe it can be computed by residue theory in complex function.
A: Your series is a Lambert series and in the case of constant coefficients $a_n$ it may be rewritten as a classical Jacobi theta function (for $z=0$) : $$\theta_3(z,q):=\sum_{n=-\infty}^\infty q^{n^2}e^{2\pi iz}$$
using the relation $(8)$ from the initial MathWorld link with $\;q=\dfrac 15$ :
$$\sum_{n=1}^\infty \dfrac {q^n}{1+q^{2n}}=\frac{\theta_3(0,q)^2-1}4$$
This is in fact an identity from Jacobi (a proof is given in chap.$9$ of Borwein&Borwein's excellent book "Pi and the AGM").
It may be written too as an elliptic function as provided in relation $(2.7)$ of this paper from Stephen Milne "Infinite families of exact sums of squares formulas, Jacobi elliptic functions, continued fractions, and Schur functions". 
Things may become more complicated instead of more simple but not less interesting !
