<s>[INST] Evaluate the value of series $$\sum^\infty_{k=0 }\frac{2^k}{5^{2^k}+1}$$
The hint is expanding it to double series.
So, $$\frac{2^k}{5^{2^k}+1}=\frac{2^k}{5^{2^k}}\frac{1}{1+5^{-2^k}}=\frac{2^k}{5^{2^k}}\sum_{n=0}^\infty(-1)^n5^{-n2^k}=\sum_{n=0}^\infty(-1)^n\frac{2^k}{5^{2^k(n+1)}}$$
Thus $$\sum^\infty_{k=0 }\frac{2^k}{5^{2^k}+1}=\sum_{k=0}^\infty\sum_{n=0}^\infty(-1)^n\frac{2^k}{5^{2^k(n+1)}}$$
Since $$\sum^\infty_{k=0 }\frac{2^k}{5^{2^k}+1}\lt \sum_{k=0}^\infty\frac{2^k}{5^k}=\frac{5}{3}$$
It is always increasing and has an upper bond. So it converges.
Therefore the order of the double series is exchangeable. But what to do next? 

 [/INST] Telescopic Solution
Notice that
$$
\frac{2^k}{5^{2^k}-1}-\frac{2^k}{5^{2^k}+1}=\frac{2^{k+1}}{5^{2^{k+1}}-1}
$$
Therefore, we get a telescoping series:
$$
\begin{align}
\sum_{k=0}^\infty\frac{2^k}{5^{2^k}+1}
&=\sum_{k=0}^\infty\left(\frac{2^k}{5^{2^k}-1}-\frac{2^{k+1}}{5^{2^{k+1}}-1}\right)\\[6pt]
&=\frac14
\end{align}
$$

Analysis of the Solution Provided

Let $\Xi=\sum_{k=0}^\infty\frac{2^k}{5^{2^k}+1}$. Note that
  $$
\frac{2^k}{5^{2^k}+1}=\frac{2^k}{5^{2^k}}\frac1{1+5^{-2^k}}=\sum_{n=0}^\infty(-1)^n\frac{2^k}{5^{2^k(n+1)}}.\tag1
$$

$(1)$ uses the sum of a geometric series

Therefore, we have
  $$
\Xi=\sum_{k=0}^\infty\sum_{n=0}^\infty(-1)^n\frac{2^k}{5^{2^k(n+1)}}
=\sum_{N=1}^\infty\frac1{5^N}\sum_{N=(n+1)2^k}(-1)^n2^k.\tag2
$$

In $(2)$, we have collected all the terms with a given power of $5$ in the denominator. Now comes the part that seems to be what is causing the difficulty:

Given any $N\in\mathbb{N}$, let $N=a2^m$ where $a$ and $m$ are non-negative integers and $a$ is odd. Then, for fixed $N$, we have
  $$
\sum_{N=(n+1)2^k}(-1)^n2^k=\sum_{k=0}^m(-1)^{a2^{m-k}-1}2^k
=2^m-\sum_{k=0}^{m-1}2^k=1.\tag3
$$

In other words, for any positive $N\in\mathbb{N}$, we can uniquely factor $N=a2^m$ where $a$ is odd. In $(2)$, we want to use all $n$ and $k$ so that $N=a2^m=(n+1)2^k$. For $0\le k\le m$, we have $n+1=a2^{m-k}$. Since $a$ is odd, $n$ is even for $k=m$ and odd for all $k\lt m$. Therefore,
$$
\sum_{N=(n+1)2^k}(-1)^n2^k=2^m-2^{m-1}-2^{m-2}-\cdots-1=1\tag4
$$
Plugging $(4)$ into $(2)$ yields

Therefore,
  $$
\Xi=\sum_{N=1}^\infty\frac1{5^N}=\frac14.\tag5
$$

</s>