<s>[INST] Compute $\sum_{k\geq0}\frac{2^k}{5^{2^k}+1}$ Problem Statement:
Compute:
$$\sum_{k\geq0}\frac{2^k}{5^{2^k}+1}$$
First we note that $\sum_{k\geq0}\frac{2^k}{5^{2^k}+1}< \sum_{k\geq0}\frac{2^k}{5^{2^k}}< \sum_{k\geq0}\frac{2^k}{5^{k}}< \infty$, so that this sum converges. We can now do the following manipulations:
\begin{align}
    \sum_{k\geq0}\frac{2^k}{5^{2^k}+1}&= \sum_{k\geq0}\frac{\frac{2^k}{5^{2^k}}}{1+ \frac{1}{5^{2^k}}}\\
    &= \sum_{k\geq0}\frac{2^k}{5^{2^k}}\left(\sum_{j\geq0}\frac{(-1)^j}{(5^{2^k})^j}\right)\quad \quad \text{geometric series with $r = -\frac{1}{5^{2^k}}$}\\
    &= \sum_{k\geq0}2^k\left(\sum_{j\geq0}\frac{(-1)^j}{(5^{2^k})^{j+1}}\right)\\
    &=\sum_{j\geq0}(-1)^j\left(\sum_{k\geq0}\frac{2^k}{(5^{j+1})^{2^k}}\right) \quad \quad \text{swap sums due to convergence}
\end{align}
I was hoping that after swapping sums the new summation would be easier to evaluate. However, this doesn't appear to be so. Any hints would be greatly appreciated.
 [/INST] This is a beautiful sum. Note:
$$\frac 1 {n-1} -\frac1{n+1} =\frac {2}{n^2-1}$$
Thus:
$$\begin{align}\frac 1 {n-1} -\frac1{n+1}-\frac {2}{n^2+1} &=\frac {2}{n^2-1}-\frac {2}{n^2+1}\\& = \frac {2^2}{n^{2^2}-1}\\\frac 1 {n-1} -\frac1{n+1}-\frac {2}{n^2+1}-\frac {2^2}{n^{2^2}+1} &= \frac {2^2}{n^{2^2}-1}-\frac {2^2}{n^{2^2}+1}\quad\quad\quad\ \  \\&= \frac {2^3}{n^{2^3}-1}\end{align}$$
Now complete the pattern.
EDIT: After a brief discussion with @Integrand, we concluded that this is much easier to explain with a telescoping sum, and by observing:
$$\frac {2^k} {n^{2^k}+1} = \frac{2^k} {n^{2^k}-1} - \frac{2^{k+1}} {n^{2^{k+1}}-1}$$
Now
$$\begin{align}\sum_{k=0}^N\frac {2^k} {n^{2^k}+1} &= \sum_{k=0}^N\left(\frac{2^k} {n^{2^k}-1} - \frac{2^{k+1}} {n^{2^{k+1}}-1}\right)\\&= \sum_{k=0}^N\frac{2^k} {n^{2^k}-1} - \sum_{k=1}^{N+1}\frac{2^{k}} {n^{2^{k}}-1}\\&=\frac{1}{n-1}-\frac{2^{N+1}}{n^{2^{N+1}}-1}\end{align}$$
and we see that $\dfrac{2^{N+1}}{n^{2^{N+1}}-1} \to 0$ as $N\to \infty$.
</s>