Find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $ how can I find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $?
It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...
Let consider sum no to infinity, but to n
 $$ \sum_{k=1}^{n} (k\cdot \ln \frac{2k+1}{2k-1} - 1) =$$
$$ ln \frac{3}{1}\cdot \left(\frac{5}{3}\right)^2 \cdot...\cdot \left(\frac{2n+1}{2n-1}\right)^n - n = ln \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1} - n $$
but $$ n = ln e^n $$
so
it will be $$ln\frac{1}{e^n} \cdot \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1}$$
So the limit of it is $-\infty$
Have I done this well or I missed sth?
 A: We have that
$$\sum_{n=1}^{N} \left(n\cdot \ln \frac{2n+1}{2n-1} - 1\right)=\sum_{n=1}^{N} \left(n\cdot \ln (2n+1)-n\ln (2n-1) - 1\right)=$$
$$=(1\cdot \ln 3-1\cdot \ln1-1)+(2\cdot \ln 5-2\cdot \ln3-1)+(3\cdot \ln 7-3\cdot \ln5-1)+\ldots=$$
$$=-\ln(3\cdot 5\cdot 7\cdot \ldots\cdot (2N-1))+N\cdot\ln(2N+1)-N=$$$$=-\ln\left(\frac{(2N)!}{2^NN!}\right)+N\cdot\ln(2N+1)-N=$$
$$=\ln\left(\frac{(2^N)^2N!N^N}{(2N)!e^N}\right)+N\cdot\ln\left(1+\frac1{2N}\right)$$
and by Stirling's approximation $N!\sim \sqrt{2\pi N}\left(\frac{N}{e}\right)^N$
$$\frac{(2^N)^2N!N^N}{(2N!)e^N}\sim\frac{(2^N)^2N^N}{e^N}\frac{\sqrt{2\pi N}}{\sqrt{4\pi N}}\frac{N^Ne^{2N}}{e^N4^NN^{2N}}=\frac{1}{\sqrt 2}$$
A: If I'm not mistaken, the actual sum is
$$ \frac{1 - \ln(2)}{2}$$
A: Let $f(x)=\displaystyle\sum_{n=1}^{\infty}\left(n\ln\frac{n+x}{n-x}-2x\right)$ for $x\in(-1,1)$. Then
$$f'(x)=\displaystyle\sum_{n=1}^{\infty}\frac{2x^2}{n^2-x^2}=1-\pi x\cot\pi x$$
(termwise differentiation is admissible because of uniform convergence of the latter series in $[-a,a]$ for any $0<a<1$; the second equality is known). Thus,
$$f(x)=x-\frac{1}{\pi}\int_{0}^{\pi x}t\cot t\,dt=x(1-\ln\sin\pi x)+\frac{1}{\pi}\int_{0}^{\pi x}\ln\sin t\,dt.$$
Your sum is $f(1/2)=(1-\ln2)/2$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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With $\ds{N \in \mathbb{N}_{\geq 1}}$:

\begin{align}
&\bbox[#ffd,10px]{\sum_{n = 1}^{N}\bracks{n\ln\pars{2n + 1 \over 2n - 1} - 1}}
\\[5mm] = &\
\sum_{n = 1}^{N}n\ln\pars{2n + 1} - \sum_{n = 1}^{N}n\ln\pars{2n - 1} - N
\\[5mm] = &\
-N + \sum_{n = 0}^{N}n\ln\pars{2n + 1} -
\sum_{n = 0}^{N - 1}\pars{n + 1}\ln\pars{2n + 1}
\\[5mm] = &\
-N + N\ln\pars{2N + 1} -N\ln\pars{2} -
\sum_{n = 0}^{N - 1}\ln\pars{n + {1 \over 2}}
\\[5mm] = &\
-N + N\ln\pars{N + {1 \over 2}} -
\ln\pars{\prod_{n = 0}^{N - 1}\bracks{n + {1 \over 2}}}
\\[5mm] = &\
-N + N\ln\pars{N + {1 \over 2}} -
\ln\pars{\bracks{N - 1/2}! \over \Gamma\pars{1/2}}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}&\
-N + N\ln\pars{N + {1 \over 2}} -
\ln\pars{\root{2\pi}\bracks{N - 1/2}^{N}\expo{-N + 1/2} \over \root{\pi}}
\\[5mm] = &\
-N + N\ln\pars{N + {1 \over 2}} -
\ln\pars{2^{1/2}N^{N}\bracks{1 - {1/2 \over N}}^{N}
\expo{-N + 1/2}}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim} &\
-N + N\ln\pars{N + {1 \over 2}} -
\bracks{{1 \over 2}\,\ln\pars{2} + N\ln\pars{N} - N}
\\[5mm] = &\
\underbrace{N\ln\pars{1 + {1 \over 2N}}}
_{\ds{\stackrel{\mrm{as}\ N\ \to\ \infty}{\to}\ {1 \over 2}}}\
-\ {1 \over 2}\,\ln\pars{2}\label{1}\tag{1}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\to} &\
\bbx{1 - \ln\pars{2} \over 2} \approx 0.1534
\end{align}
