TITLE: Summation of an infinite Exponential series
QUESTION [15 upvotes]: 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

REPLY [22 votes]: 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 $n \to \infty$.