Evaluate $\sum \frac{2^k}{3^{2^k}+1}$

Question

Evaluate the sum $$\sum_{k=0}^{\infty} \left ( \frac{2^k}{3^{2^k}+1} \right )$$

I first tried to see weather the sum actually converges, by ratio test it does as $\lim_{n\to \infty}|\frac{a_{n+1}}{a_n}|<1$ where $a_n$ is the nth term. Now I am left with finding the actual sum. I thought to write it so that it can telescope, but can't find any suitable representations. Any hints for the evaluation part? Any Taylor polynomials etc?

Update: Wolfram says its $1/2$

Numerical evidence seems to suggest $1/2$, and this seems reasonable to me. I'll see if I can come up with a way to prove it. — A. Thomas Yerger, May 29 '17 at 17:43
Look at the @RonGordon 's answer here: A closed form for the sum $S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1}$ is ... — Sil, May 29 '17 at 18:02
I got it doing summation by parts! But I made a sign error, so I'm going to find that error before I post. — A. Thomas Yerger, May 29 '17 at 18:03
I will be very much interested in seeing use of Abel summation, I haven't seen it in action a lot. — , May 29 '17 at 18:04
@Glasgow39 I deleted my post, as I think something went pretty wrong. I'm quite bad at arithmetic it seems. — A. Thomas Yerger, May 29 '17 at 19:03

score 1 · Accepted Answer · answered May 29 '17 at 19:26

1

$$\frac{2^k}{3^{2^k}+1}=\frac{2^k}{3^{2^k}-1}-\frac{2^{k+1}}{3^{2^{k+1}}-1}$$ So telescoping is the way to go, here.

answered May 29 '17 at 19:26

score 0 · Answer 2 · answered May 29 '17 at 22:01

I'm going to rederive your telescoping of $\dfrac{2^k}{3^{2^k}+1} =\dfrac{2^k}{3^{2^k}-1}-\dfrac{2^{k+1}}{3^{2^{k+1}}-1} $ and see if I can generalize it.

$\begin{array}\\ \dfrac{2^k}{3^{2^k}-1}-\dfrac{2^{k+1}}{3^{2^{k+1}}-1} &=\dfrac{2^k(3^{2^{k+1}}-1)-2^{k+1}(3^{2^k}-1)}{(3^{2^k}-1)(3^{2^{k+1}}-1)}\\ &=\dfrac{2^k(3^{2^{k+1}}-1-2(3^{2^k}-1)}{(3^{2^k}-1)(3^{2^{k+1}}-1)}\\ &=\dfrac{2^k((3^{2^{k}})^2-2(3^{2^k}+1)}{(3^{2^k}-1)(3^{2^{k+1}}-1)}\\ &=\dfrac{2^k(3^{2^{k}}-1)^2}{(3^{2^k}-1)(3^{2^{k+1}}-1)}\\ &=\dfrac{2^k(3^{2^{k}}-1)}{(3^{2^{k+1}}-1)}\\ &=\dfrac{2^k(3^{2^{k}}-1)}{(3^{2^{k+1}})^2-1}\\ &=\dfrac{2^k(3^{2^{k}}-1)}{(3^{2^{k}}-1)(3^{2^{k}}+1)}\\ &=\dfrac{2^k}{3^{2^{k}}+1}\\ \end{array} $

It looks like the "3" can be anything, while the "2" is essential since it is used in $x^2-1 = (x-1)(x+1)$. So, replacing $3$ by $m$, this becomes

$\begin{array}\\ \dfrac{2^k}{m^{2^k}-1}-\dfrac{2^{k+1}}{m^{2^{k+1}}-1} &=\dfrac{2^k(m^{2^{k+1}}-1)-2^{k+1}(m^{2^k}-1)}{(m^{2^k}-1)(m^{2^{k+1}}-1)}\\ &=\dfrac{2^k(m^{2^{k+1}}-1-2(m^{2^k}-1)}{(m^{2^k}-1)(m^{2^{k+1}}-1)}\\ &=\dfrac{2^k((m^{2^{k}})^2-2(m^{2^k}+1)}{(m^{2^k}-1)(m^{2^{k+1}}-1)}\\ &=\dfrac{2^k(m^{2^{k}}-1)^2}{(m^{2^k}-1)(m^{2^{k+1}}-1)}\\ &=\dfrac{2^k(m^{2^{k}}-1)}{m^{2^{k+1}}-1}\\ &=\dfrac{2^k(m^{2^{k}}-1)}{(m^{2^{k+1}})^2-1}\\ &=\dfrac{2^k(m^{2^{k}}-1)}{(m^{2^{k}}-1)(m^{2^{k}}+1)}\\ &=\dfrac{2^k}{m^{2^{k}}+1}\\ \end{array} $

From this, $\sum_{k=0}^{\infty} \dfrac{2^k}{m^{2^{k}}+1} =\sum_{k=0}^{\infty} \left(\dfrac{2^k}{m^{2^k}-1}-\dfrac{2^{k+1}}{m^{2^{k+1}}-1} \right) =\dfrac1{m-1} $.

For $m=3$ we get OP's result.

Evaluate $\sum \frac{2^k}{3^{2^k}+1}$

2 Answers2