According to the accepted answer of this question, we can apply Euler's series transformation to derive that $$\displaystyle{\sum_{n=1}^{\infty}\frac{1}{n^2}=\sum_{n=1}^{\infty}\frac{3}{n^2\binom{2n}{n}}}.$$ I am wondering, how can we do that? Can you provide me an explanation?
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A proof by pure creative telescoping has been linked in the comments above, but there also is an interesting proof that comes from manipulations of a logarithmic integral.
If we set $$ I = -\int_{0}^{\pi/2}\log\left(1-\frac{1}{4}\sin^2 x\right)\frac{dx}{\sin x}$$ by expanding $-\log\left(1-\frac{1}{4}\sin^2 x\right)$ as a Taylor series in $\sin x$ we get that $$\begin{eqnarray*} \color{blue}{I} = \sum_{n\geq 1}\frac{1}{n 4^n}\int_{0}^{\pi/2}\sin(x)^{2n-1}\,dx = \sum_{n\geq 1}\frac{1}{4n(2n-1)\binom{2n-2}{n-1}}=\color{blue}{ \frac{1}{2}\sum_{n\geq 1}\frac{1}{n^2\binom{2n}{n}}} \end{eqnarray*}$$ and by applying the tangent half-angle substitution we get that: $$ I = \int_{0}^{1}-\log\left(1-\left(\frac{t}{1+t^2}\right)^2\right)\frac{dt}{t} $$ where the rational function $1-\left(\frac{t}{1+t^2}\right)^2$ can be expressed through products and ratios of polynomials of the form$1-t^m$. Eureka, since: $$ I_m=-\int_{0}^{1}\frac{\log(1-t^m)}{t} = \sum_{n\geq 1}\frac{1}{n}\int_{0}^{1}t^{mn-1}\,dt = \sum_{n\geq 1}\frac{1}{mn^2}=-\frac{\zeta(2)}{m}$$ implies: $$ \color{blue}{I} = \left(I_2-2 I_4+I_6\right)=\color{blue}{\frac{1}{6}\zeta(2)}$$ as wanted.
Since we have $$ 2\arcsin^2(x) = \sum_{n\geq 1}\frac{(2x)^{2n}}{n^2\binom{2n}{n}} $$ as proved here, the acceleration formula implies that $\zeta(2)=6\arcsin^2\left(\frac{1}{2}\right)=\frac{\pi^2}{6}$.
Jack D'Aurizio
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\displaystylein titles. – Did Jun 23 '17 at 16:05