Central Binomial coefficient series and zeta(2)

In this entry we will prove the following result that appears in this Twitter post

$\begin{align*} \zeta(2)=\frac23\sum_{n=1}^\infty \frac{\binom{2n}{n}}{ 2^{2n} n}\sum_{k=0}^{n-1}\frac{1}{2k+1} \label{1} \end{align*}$

Click here for the proof.

In the proof we used the previous result

$\begin{align*} \sum_{n=1}^\infty \binom{2n}{n}\frac{x^n}{ 2^{2n} n}&=2\ln\left(\frac{2}{1+\sqrt{1-x}} \right) \end{align*}$

proved here.

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