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.

  • $\begingroup$ 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. $\endgroup$ – davidlowryduda Apr 17 '18 at 17:14
  • $\begingroup$ Euler-Maclaurin may be the way $\endgroup$ – Pierpaolo Vivo Apr 17 '18 at 18:35
  • $\begingroup$ @PierpaoloVivo I don't see how this question relates to it $\endgroup$ – Rohan Shinde Apr 17 '18 at 18:40
  • 4
    $\begingroup$ Numerically, it "seems to go" to $\displaystyle\color{red}{1 \over 4}$. $\endgroup$ – Felix Marin Apr 17 '18 at 19:14
  • $\begingroup$ @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 $\endgroup$ – Rohan Shinde Apr 18 '18 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 $a$ $$\lim_{n\to \infty} \left(\sum_{r=0}^n \frac {2^r}{a^{2^r}+1}\right) =\frac {1}{a-1}$$

  • 2
    $\begingroup$ +1. Very clever. $\endgroup$ – Felix Marin Apr 18 '18 at 16:40
  • 1
    $\begingroup$ @Felix Marin Thanks a lot $\endgroup$ – Rohan Shinde Apr 18 '18 at 18:39
  • 2
    $\begingroup$ Good job, +1) . $\endgroup$ – xpaul Apr 18 '18 at 20:52
  • 2
    $\begingroup$ Telescoping sums are so nice. (+1) $\endgroup$ – Kirk Fox Apr 19 '18 at 1:22
  • 1
    $\begingroup$ @xpaul Thanks... $\endgroup$ – Rohan Shinde Apr 19 '18 at 3:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.