# Prove that $-2\log(2) = -2 + \sum_{n=1}^{\infty}\frac{1}{n(2n+1)}$

How to prove that

$$-2\log(2) = -2 + \sum_{n=1}^{\infty}\frac{1}{n(2n+1)}$$

I know that this sum is equal to $\phi(1/2)+\gamma$ where $\phi(x)$ is the digamma function and $\gamma$ is the Euler-Mascheroni constant, but I want to evaluate the sum without knowing it.

$$\frac{1}{2n(2n+1)}=\frac1{2n}-\frac1{2n+1}.$$ $$\sum_{n=1}^\infty\frac1{n(2n+1)} =2\sum_{n=1}^\infty\left(\frac1{2n}-\frac1{2n+1}\right) =2\sum_{m=2}^\infty\frac{(-1)^m}{m}.$$ Of course, $$\log2=-\sum_{m=1}^\infty\frac{(-1)^m}{m}.$$
• @Alex make sure to tick this as an answer :) $\ \ \color{green}{\checkmark}$ – Mr Pie Feb 17 '18 at 8:10
Or you can prove forward: \begin{align}-2\ln2= &-2\left(1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots\right)=\\ &-2\left(1-\frac{1}{2\cdot 3}-\frac{1}{4\cdot 5}-\cdots\right)=\\ &-2\left(1-\sum_{n=1}^{\infty} \frac{1}{2n\cdot (2n+1)}\right)=\\ &-2+\sum_{n=1}^{\infty} \frac{1}{n(2n+1)}.\end{align}