Evaluate $\lim_{n\to \infty} \left( \sum_{r=0}^n \frac {2^r}{5^{2^r}+1}\right) $
$$\lim_{n\to \infty} \left( \sum_{r=0}^n \frac {2^r}{5^{2^r}+1}\right) $$ $$=\lim_{n\to \infty} \sum_{r=0}^n\left( \frac {2^r}{5^{2^r}+1}\cdot \frac {5^{2^r}-1}{5^{2^r}-1}\right) $$ $$=\lim_{n\to \infty} \sum_{r=0}^n \left(\frac {2^r((5^{2^r}+1)-2)}{5^{2^{r+1}}-1}\right) $$ $$=\lim_{n\to \infty} \sum_{r=0}^n \left( \frac {2^r}{5^{2^r}-1} -\frac {2^{r+1}}{5^{2^{r+1}}-1}\right)$$ $$=\frac {1}{5-1}=\frac 14$$
Note: In fact using this method it can be proved that for any natural number $a$ (except of course $a=1$) $$\lim_{n\to \infty} \left(\sum_{r=0}^n \frac {2^r}{a^{2^r}+1}\right) =\frac {1}{a-1}$$