Summation of an infinite Exponential series

Solution 1:

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

$$\sum_{r=0}^\infty \frac{2^r}{5^{2^r} + 1} = \sum_{r=0}^\infty \frac{2^r\left(\frac15\right)^{2^r}}{1 + \left(\frac15\right)^{2^r}} = \frac{\frac15}{1-\frac15} = \frac14$$

Solution 2:

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 $n \to \infty$.