Does anybody know how to prove this series?
$$\sum_{n=1}^{\infty} \left( n\log \left(\frac{2n+1}{2n-1}\right)-1\right) = \frac{1-\log 2}{2}$$
I arrived at this through Mathematica.
I tried writing $\log \left(\frac{2n+1}{2n-1}\right)$ as $\int_0^1 \frac{1}{\frac{2n-1}{2}+x}dx$ and $-\sum_{k=1}^\infty \frac{(-2)^k}{k(2n-1)^k}$ but none of them worked.
Note that
\begin{align*} \sum_{n=1}^{\infty}\left(n \log \left(\frac{2n+1}{2n-1}\right) - 1 \right) &=\lim_{N\to\infty} \sum_{n=1}^{N}\left(n \log \left(\frac{2n+1}{2n-1}\right) - 1 \right)\\ &=\lim_{N\to\infty} \log\left[ e^{-N} \prod_{n=1}^{N} \left(\frac{2n+1}{2n-1}\right)^{n} \right] \\ &=\lim_{N\to\infty} \log\left[ e^{-N} \frac{2^{N} N! (2N+1)^{N}}{(2N)!} \right]. \end{align*}
By Stirling's formula, it follows that
$$ e^{-N} \frac{2^{N} N! (2N+1)^{N}}{(2N)!} \sim \sqrt{\frac{e}{2}}. $$
This immediately yields the desired answer.
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{n = 1}^{\infty}\bracks{n\ \ln\pars{2n + 1 \over 2n - 1} - 1} ={1 - \ln\pars{2} \over 2}}$
With ${\ds{\quad N \in {\mathbb N}\,,\quad N \geq 1}}$: \begin{align}&\color{#c00000}{\sum_{n = 1}^{N}n\ \ln\pars{2n + 1 \over 2n - 1}} =\sum_{n = 1}^{N}n\int_{-1}^{1}{\dd x \over x + 2n} =\half\int_{0}^{1}\sum_{n = 1}^{N}\pars{{n \over n + x/2} + {n \over n - x/2}} \,\dd x \\[3mm]&=\half\int_{0}^{1}\sum_{n = 1}^{N} \pars{1 - {x/2 \over n + x/2} + 1 + {x/2 \over n - x/2}}\,\dd x \\[3mm]&=N + {1 \over 4}\int_{0}^{1} x\sum_{n = 1}^{N}{x \over \pars{n + x/2}\pars{n - x/2}}\,\dd x \\[3mm]&=N + {1 \over 4}\int_{0}^{1} x\sum_{n = 0}^{N - 1}{x \over \pars{n + 1 + x/2}\pars{n + 1 - x/2}}\,\dd x \end{align}
$$ \color{#c00000}{\sum_{n = 1}^{N}\bracks{n\ln\pars{2n + 1 \over 2n - 1} - 1}} ={1 \over 4}\int_{0}^{1} x\sum_{n = 0}^{N - 1}{x \over \pars{n + 1 + x/2}\pars{n + 1 - x/2}}\,\dd x $$
\begin{align} &\color{#c00000}{\sum_{n = 1}^{\infty}\bracks{n\ln\pars{2n + 1 \over 2n - 1} - 1}} ={1 \over 4}\int_{0}^{1}x\bracks{% \Psi\pars{1 + {x \over 2}} - \Psi\pars{1 - {x \over 2}}}\,\dd x \\[3mm]&={1 \over 4}\int_{0}^{1}x\bracks{% {2 \over x} + \Psi\pars{x \over 2} - \Psi\pars{1 - {x \over 2}}}\,\dd x =\half - {1 \over 4}\int_{0}^{1}x\pi\cot\pars{\pi x \over 2}\,\dd x \end{align}
$$ \color{#c00000}{\sum_{n = 1}^{\infty}\bracks{n\ln\pars{2n + 1 \over 2n - 1} - 1}} =\half - {1 \over \pi}\color{#00f}{\int_{0}^{\pi/2}x\cot\pars{x}\,\dd x}\tag{1} $$
\begin{align}&\color{#00f}{\int_{0}^{\pi/2}x\cot\pars{x}\,\dd x} =\left.x\ln\pars{\sin\pars{x}}\vphantom{\LARGE A}\right\vert_{0}^{\pi/2} -\ \underbrace{\int_{0}^{\pi/2}\ln\pars{\sin\pars{x}}\,\dd x} _{\ds{-\,{\pi\ln\pars{2} \over 2}}}\ =\ \color{#00f}{{\pi\ln\pars{2} \over 2}} \end{align} The $\ds{\large\ln\pars{\sin\pars{\cdots}}}$-integral is reported by M.SE as a frequent question. Replacing the last result in $\pars{1}$:
$$\color{#66f}{\large% \sum_{n = 1}^{\infty}\bracks{n\ \ln\pars{2n + 1 \over 2n - 1} - 1} ={1 - \ln\pars{2} \over 2}} \approx 0.1534 $$
$$n \log \left(\dfrac{2n+1}{2n-1}\right) = n \left(\log(1+1/2n) - \log(1-1/2n)\right)$$ \begin{align} \log(1+x) - \log(1-x) & = \left(x - \dfrac{x^2}2 + \dfrac{x^3}3 \mp \cdots\right) - \left(-x - \dfrac{x^2}2 - \dfrac{x^3}3 - \cdots\right)\\ & = 2\left(x + \dfrac{x^3}3 + \dfrac{x^5}5 + \cdots \right) \end{align} Hence, $$n \left(\log(1+1/2n) - \log(1-1/2n)\right) = 2n \left(\sum_{k=0}^{\infty} \dfrac1{(2k+1)(2n)^{2k+1}}\right) = \sum_{k=0}^{\infty} \dfrac1{(2k+1)(2n)^{2k}}$$ Hence, $$n \log \left(\dfrac{2n+1}{2n-1}\right) - 1 = \sum_{k=1}^{\infty} \dfrac1{(2k+1)(2n)^{2k}}$$ $$\sum_{n=1}^{\infty} \left(n \log \left(\dfrac{2n+1}{2n-1}\right) - 1\right) = \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \dfrac1{(2k+1)(2n)^{2k}} = \sum_{k=1}^{\infty} \dfrac{\zeta(2k)}{(2k+1) \cdot 2^{2k}}$$
\begin{align}f(m) & = \sum_{n=1}^m \left(n \log\left(\dfrac{2n+1}{2n-1}\right)-1\right)\\ & = \left( 1\log(3) - 1 + 2 \log(5) - 2 \log(3) + 1 + \cdots \right)\\ & = m - \log(3) - \log(5) - \cdots - \log(2m-1) + m \log(2m+1) \end{align} Now $$\log(1) + \log(3) + \log(5) + \cdots + \log(2m-1) = \log \left(\dfrac{(2m)!}{2^m \cdot m!}\right)$$ Hence, $$f(m) = m + m \log(2m+1) - \log \left(\dfrac{(2m)!}{2^m \cdot m!}\right)$$ Now use Stirling to get your answer.
