# 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.

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 Try $\log((2n+1)/(2n-1))=\log(2n+1)-\log(2n-1)$ ? – ᴊ ᴀ s ᴏ ɴ Mar 17 at 5:45

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.

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$$n \log \left(\dfrac{2n+1}{2n-1}\right) = n \left(\log(1+1/2n) - \log(1-1/2n)\right)$$ \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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