<s>[INST] 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?
 [/INST] 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 $f(1/2)=(1-\ln2)/2$.
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