Summation of an infinite Exponential series Q. Find the value of - $$ \lim_{n\to\infty} \sum_{r=0}^{n} \frac{2^r}{5^{2^r} +1} $$
My attempt - I seem to be clueless to this problem. Though I think that later terms would be much small and negligible( Imagine how large would be $ 5^{2^r} $ after 3-4 terms), so I calculated the sum of first 3-4 terms and my answer was just around the actual answer    ( just a difference of $0.002$ ). But I wonder if there is an actual method to solve this problem? If it is, would you please share it to me? 
Any help would be appreciated
 A: Recall every natural number $m$ can be uniquely decomposed into a sum of powers of $2$,
$$m = \sum_{r=0}^\infty b_r 2^r\quad\text{ where }\quad b_r \in \{ 0, 1 \}$$
For any $x \in (0,1)$, this leads to
$$\prod_{r=0}^\infty \left(1 + x^{2^r}\right) = \sum_{m=0}^\infty  x^m = \frac{1}{1-x}$$
Taking logarithm on both sides and apply $x\frac{d}{dx}$ to them, we obtain
$$\sum_{r=0}^\infty \frac{2^r x^{2^r}}{1 + x^{2^r}} = \frac{x}{1-x}$$
Substitute $x$ by $\frac15$, we find
$$\sum_{r=0}^\infty \frac{2^r}{5^{2^r} + 1} = \sum_{r=0}^\infty
\frac{2^r\left(\frac15\right)^{2^r}}{1 + \left(\frac15\right)^{2^r}}
= \frac{\frac15}{1-\frac15} = \frac14$$
A: Another user pisco originally posted this, but kept deleting his/her answer. It however is completely elementary and hence in my opinion superior to using power series and termwise differentiation (which require non-trivial theorems):
$$\frac{2^r}{5^{2^r}+1} = \frac{2^r}{5^{2^r}-1} - \frac{2^{r+1}}{5^{2^{r+1}}-1}.$$
Then just take $\displaystyle\sum_{r=0}^n$ of both sides and observe that the 'remainder' term vanishes as $n \to \infty$.
