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

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.

• Why do you think this series should have a nice evaluation? An extremely accurate evaluation will come from just the first few terms, since this series converges very, very rapidly. Apr 17, 2018 at 17:14
• Euler-Maclaurin may be the way Apr 17, 2018 at 18:35
• @PierpaoloVivo I don't see how this question relates to it Apr 17, 2018 at 18:40
• Numerically, it "seems to go" to $\displaystyle\color{red}{1 \over 4}$. Apr 17, 2018 at 19:14
• @FelixMarin Yeah I got to know about that from Wolfy but I want an algebraic hint or some kind method to do it by hand by proper means Apr 18, 2018 at 1:35

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