<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] 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 $N!\sim \sqrt{2\pi N}\left(\frac{N}{e}\right)^N$
$$\frac{(2^N)^2N!N^N}{(2N!)e^N}\sim\frac{(2^N)^2N^N}{e^N}\frac{\sqrt{2\pi N}}{\sqrt{4\pi N}}\frac{N^Ne^{2N}}{e^N4^NN^{2N}}=\frac{1}{\sqrt 2}$$
</s>