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.
Best Answer
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.