<s>[INST] Series involving log $\sum_{n=1}^{\infty} \left( n\log \left(\frac{2n+1}{2n-1}\right)-1\right) = \frac{1-\log 2}{2}$ 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.
 [/INST] $$n \log \left(\dfrac{2n+1}{2n-1}\right) = n \left(\log(1+1/2n) - \log(1-1/2n)\right)$$
\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.
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