How to prove $\ln{6}=\sum_{n=1}^{\infty}\sum_{r=2}^{\infty}\left({1\over r^{2n}}+{2\over (r+1)^{2n}}+{1\over (r+2)^{2n}}\right)$?

Question

I need help, on how to prove

$$\ln{6}=\sum_{n=1}^{\infty}\sum_{r=2}^{\infty}\left({1\over r^{2n}}+{2\over (r+1)^{2n}}+{1\over (r+2)^{2n}}\right).$$

Any hints?

Answer 1

I don't find the announced result.

Hint. The double series is absolutely convergent, then one may interchange the summations, using the geometric standard evaluation, $$ \sum_{n=1}^{\infty}\left({1\over r^{2n}}+{2\over (r+1)^{2n}}+{1\over (r+2)^{2n}}\right)=\frac{1}{2 (r-1)}+\frac{1}{r}-\frac{1}{r+2}-\frac{1}{2 (r+3)} $$ then one may observe that terms telescope to get

$$ \sum_{n=1}^{\infty}\sum_{r=2}^{\infty}\!\left(\!{1\over r^{2n}}\!+\!{2\over (r+1)^{2n}}\!+\!{1\over (r+2)^{2n}}\!\right)\!=\!\sum_{r=2}^{\infty}\!\left(\!\frac{1}{2 (r-1)}\!-\!\frac{1}{2 (r+3)}\!+\!\frac{1}{r}\!-\!\frac{1}{r+2}\!\right)\!=\!\frac{15}8. $$

Answer 2

Another approach is the following one: $$ \sum_{r\geq 2}\left(\frac{1}{r^{2n}}+\frac{2}{(r+1)^{2n}}+\frac{1}{(r+2)^{2n}}\right) = 4(\zeta(2n)-1)-\frac{3}{2^{2n}}-\frac{1}{3^{2n}}\tag{1}$$ giving: $$ \sum_{n\geq 1}\sum_{r\geq 2}\left(\frac{1}{r^{2n}}+\frac{2}{(r+1)^{2n}}+\frac{1}{(r+2)^{2n}}\right)=-\frac{9}{8}+4\sum_{n\geq 1}\left(\zeta(2n)-1\right)\tag{2}$$ where: $$ \sum_{n\geq 1}\left(\zeta(2n)-1\right)=\sum_{n\geq 1}\int_{0}^{+\infty}\frac{x^{2n-1}}{(2n-1)!}\frac{dx}{e^{2x}-e^{x}}=\int_{0}^{+\infty}\frac{\sinh(x)}{e^{2x}-e^{x}}\,dx\tag{3} $$ and: $$ \int_{0}^{+\infty}\frac{\sinh(x)}{e^{2x}-e^{x}}\,dx =\frac{1}{2}\int_{0}^{+\infty}\frac{e^{2x}-1}{e^{3x}-e^{2x}}\,dx = \frac{1}{2}\int_{0}^{+\infty}\frac{e^{x}+1}{e^{2x}}\,dx = \frac{3}{4}\tag{4} $$ leads to:

$$ \sum_{r\geq 2}\left(\frac{1}{r^{2n}}+\frac{2}{(r+1)^{2n}}+\frac{1}{(r+2)^{2n}}\right) = 3-\frac{9}{8}=\color{red}{\frac{15}{8}}\tag{5}.$$

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\color{#f00}{\sum_{n = 1}^{\infty}\sum_{r = 2}^{\infty}\bracks{% {1 \over r^{2n}} + {2 \over \pars{r + 1}^{2n}} + {1 \over \pars{r + 2}^{2n}}}} \\[4mm] = &\ \sum_{n = 1}^{\infty}\braces{% \sum_{r = 2}^{\infty} \bracks{{1 \over r^{2n}} + {1 \over \pars{r + 1}^{2n}}} + \sum_{r = 3}^{\infty}\bracks{{1 \over r^{2n}} + {1 \over \pars{r + 1}^{2n}}}} \\[4mm] = &\ \sum_{n = 1}^{\infty}\bracks{% {1 \over 2^{2n}} + {1 \over 3^{2n}} + 2\sum_{r = 3}^{\infty}{1 \over r^{2n}} + 2\sum_{r = 4}^{\infty}{1 \over r^{2n}}} = \sum_{n = 1}^{\infty}\bracks{% {1 \over 2^{2n}} + {1 \over 3^{2n - 1}} + 4\sum_{r = 4}^{\infty}{1 \over r^{2n}}} \\[4mm] = &\ {1/4 \over 1 - 1/4} + {1/3 \over 1 - 1/9} + 4\sum_{r = 4}^{\infty}{1/r^{2} \over 1 - 1/r^{2}} = {1 \over 3} + {3 \over 8} + 2\sum_{r = 4}^{\infty}\pars{{1 \over r - 1} - {1 \over r + 1}} \\[4mm] = &\ {17 \over 24} + 2\ \underbrace{\sum_{r = 4}^{\infty}\pars{{1 \over r - 1} - {1 \over r}}} _{\ds{1 \over 3}}\ +\ 2\ \underbrace{\sum_{r = 4}^{\infty}\pars{{1 \over r} - {1 \over r + 1}}} _{\ds{1 \over 4}} ={17 \over 24} + 2\pars{{1 \over 3} + {1 \over 4}} \\[4mm] = & \color{#f00}{15 \over 8} = 1.875 \not= \fbox{$\ds{\ \ln\pars{6} \approx 1.7918\ }$}\ !!!. \end{align}