Evaluate $\lim_{n\to \infty} \left( \sum_{r=0}^n \frac {2^r}{5^{2^r}+1}\right) $

Question

Evaluate $$\lim_{n\to \infty} \left( \sum_{r=0}^n \frac {2^r}{5^{2^r}+1}\right) $$

I tried to create some infinite GP within the summation, some algebraic manipulations like adding the first and last terms of the summation to find any series popping out of it and also tried writing it in the exponential form like $5^{2^r}=e^{2^r\ln 5}$ and also tried to do some power series thing. I also tried to find any method using integrals and Riemann sums but couldn't do so.

Any hints would be greatly appreciated.

Answer 1

$$\lim_{n\to \infty} \left( \sum_{r=0}^n \frac {2^r}{5^{2^r}+1}\right) $$ $$=\lim_{n\to \infty} \sum_{r=0}^n\left( \frac {2^r}{5^{2^r}+1}\cdot \frac {5^{2^r}-1}{5^{2^r}-1}\right) $$ $$=\lim_{n\to \infty} \sum_{r=0}^n \left(\frac {2^r((5^{2^r}+1)-2)}{5^{2^{r+1}}-1}\right) $$ $$=\lim_{n\to \infty} \sum_{r=0}^n \left( \frac {2^r}{5^{2^r}-1} -\frac {2^{r+1}}{5^{2^{r+1}}-1}\right)$$ $$=\frac {1}{5-1}=\frac 14$$

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

Answer 2

The key is to understand that if $|x|<1$ then $$(1+x)(1+x^2)(1+x^4)\dots(1+x^{2^n})\dots=\frac{1}{1-x}$$ Now taking logs and differentiating it we get $$\sum_{n=0}^{\infty} \frac{2^nx^{2^n-1}}{1+x^{2^n}}=\frac{1}{1-x}$$ Putting $x=1/t,|t|>1$ we get $$\sum_{n=0}^{\infty} \frac{2^n}{1+t^{2^n}}=\frac{1}{t-1}$$ Putting $t=5$ we get the desired limit in question.