kuntungw3

2022-01-21

Find $\sum _{n=0}^{\mathrm{\infty }}\frac{{2}^{n}}{{3}^{{2}^{n}-1}+1}$

trnovitom06

Start with following infinite product expansion which is valid for $|q|<1$
$\prod _{n=0}^{\mathrm{\infty }}\left(1+{q}^{{2}^{n}}\right)=\sum _{n=0}^{\mathrm{\infty }}{q}^{n}=\frac{1}{1-q}$
Taking logarithm and apply $q\frac{d}{dq}$ on both sides, one get
$\sum _{n=0}^{\mathrm{\infty }}\frac{{2}^{n}{q}^{{2}^{n}}}{1+{q}^{{2}^{n}}}=\frac{q}{1-q}$
Substitute q by $\frac{1}{\sqrt{3}}$, one obtain
$\sum _{n=0}^{\mathrm{\infty }}\frac{{2}^{n}}{{3}^{{2}^{n-1}}+1}=\frac{1}{\sqrt{3}-1}=\frac{\sqrt{3}+1}{2}=1+\frac{\sqrt{3}-1}{2}=1+\frac{1}{\sqrt{3}+1}$
the expression you expected.

ocretz56

Theres

RizerMix

There is no reason for the oddball $\frac{{2}^{0}}{{3}^{{2}^{0-1}}}$ term so I omit it. $\sum _{r=1}^{\mathrm{\infty }}\frac{{2}^{r}}{{3}^{{2}^{r-1}}}\frac{{2}^{r}}{{3}^{{2}^{r-1}}}=\sum _{r=1}^{\mathrm{\infty }}{2}^{r}\sum _{m=1}^{\mathrm{\infty }}\left(-1{\right)}^{m-1}{3}^{-{2}^{r-1}m}$ $=\sum _{n=1}^{\mathrm{\infty }}{3}^{-n}\sum _{{2}^{r-1}|n}{2}^{r}\left(-1{\right)}^{n/{2}^{r}-1}$ $=\sum _{n=1}^{\mathrm{\infty }}{3}^{-n}\cdot 2$ $=2\frac{1/3}{1-1/3}$ $=1$ Note, given $v:={v}_{2}\left(n\right),\sum _{{2}^{r-1}|n}{2}^{r}\left(-1{\right)}^{n/{2}^{r}-1}$ is 2 if $v=0$ (i.e. n is odd) and otherwise $=-2\sum _{r=0}^{v}{2}^{r}\left(-1{\right)}^{n/{2}^{r}}=-2\left[\sum _{r=0}^{v-1}{2}^{r}-{2}^{v}\right]$ $=-2\left[\frac{{2}^{v}-1}{2-1}-{2}^{v}\right]=-2\left(-1\right)=2$ if $v>0$