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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.

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$$\frac{1}{2n(2n+1)}=\frac1{2n}-\frac1{2n+1}.$$ $$\sum_{n=1}^\infty\frac1{n(2n+1)} =2\sum_{n=1}^\infty\left(\frac1{2n}-\frac1{2n+1}\right) =2\sum_{m=2}^\infty\frac{(-1)^m}{m}.$$ Of course, $$\log2=-\sum_{m=1}^\infty\frac{(-1)^m}{m}.$$

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  • $\begingroup$ Thanks al lot:) $\endgroup$ – Alex Feb 17 '18 at 8:00
  • $\begingroup$ @Alex make sure to tick this as an answer :) $\ \ \color{green}{\checkmark}$ $\endgroup$ – Mr Pie Feb 17 '18 at 8:10
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Or you can prove forward: $$\begin{align}-2\ln2= &-2\left(1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots\right)=\\ &-2\left(1-\frac{1}{2\cdot 3}-\frac{1}{4\cdot 5}-\cdots\right)=\\ &-2\left(1-\sum_{n=1}^{\infty} \frac{1}{2n\cdot (2n+1)}\right)=\\ &-2+\sum_{n=1}^{\infty} \frac{1}{n(2n+1)}.\end{align}$$

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