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$$\sum_{n=1}^{\infty}\frac1{n^2}=\sum_{n=1}^{\infty}\frac3{n^2\binom{2n}n}\tag1$$

Note that $(1)$ is true since the LHS equals $\zeta(2)$ whereas the RHS equals $6\arcsin^2 1$ which both turn out to equal $\pi^2/6$ as is well-known. However I am little bit curious about how to show $(1)$ without actually evaluating both series.

I am aware of an elegant approach contributed by Markus Scheuer as an answer to Different methods to compute Basel problem. Although this answers my question partially I am looking for different attemps. For example within Jack D'Aurizio's notes there is a way proposed exploiting creative telescoping $($see page $5$f.$)$ which I am sadly speaking not able to understand completely yet. Hence I have come across a proof of a similiar equality concerning $\zeta(3)$ on AoPS given by pprime I am confident that there are in fact other possible methods.

I would like to see attempts of proving $(1)$ beside the mentioned which do not rely on actually showing that they both equal $\pi^2/6$. Preferably these should be in the spirit of Markus Scheuer's or Jack D'Aurizio's approaches rather than the one similiar by pprime.

EDIT I

I have found another interesting approach, again by Jack D'Aurizio, which can be found here utilizing harmonic sums and creative telescoping in combination.

EDIT II

As pointed out by Zacky on page $31$ of Jack's notes another methode can be found which makes it three possibilities provided by Jack alone. Quite impressive!

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  • $\begingroup$ Here is a similar type of series (just 2 hours ago, also answered by yourself), using $\arcsin$. But yes, here is the series evaluated directly. Seems pretty good to me. $\endgroup$ – Dietrich Burde Dec 24 '18 at 12:53
  • $\begingroup$ @DietrichBurde I am not sure how this is of use. Could you elaborate on the utility of the linked post in order to answer my question? Furthermore I am aware of this post since I posted an answer there $2$ hours ago ^^ $\endgroup$ – mrtaurho Dec 24 '18 at 12:59
  • $\begingroup$ I thought, it would be the best to show that both sides are equal to $\zeta(2)$, but you want a different solution (and I just do not know why this should be better or more interesting, but this is of course due to my missing understanding). $\endgroup$ – Dietrich Burde Dec 24 '18 at 13:08
  • $\begingroup$ @DietrichBurde Ah okay. I have not considered it from this point of view. Regarding to the value of alternative proofs: I am just interested in different approaches. For sure it is quite convincing to show both series equal the same value but out of experience I have observed that is most likely to be a way more difficult to show the equality of the series all by themselves $($see for example here: math.stackexchange.com/q/2942630 where an elegant trick was needed to show that the integrals are equivalent$)$. $\endgroup$ – mrtaurho Dec 24 '18 at 13:10
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Here is a way, although I am not sure that this is what you seek to find.

Since we have $$ \frac{1}{\binom{2n}{n}} =\frac{n!n!}{(2n)!} $$ by the binomial identity, we obtain $$ \frac{1}{n^2\binom{2n}{n}} =\frac{(n-1)!(n-1)!}{(2n)!} =\frac{\color{purple}{\Gamma(n)\Gamma(n)}}{2n\color{purple}{\Gamma(n+n)}}=\frac{1}{2n}\color{purple}{B(n,n)} $$ Therefore we get \begin{align*} \sum_{n=1}^\infty \frac{1}{n^2\binom{2n}{n}} & =\frac12\sum_{n=1}^\infty \frac{1}{n}\color{purple}{B(n,n)} =\frac12\sum_{n=1}^\infty \frac{1}{n}\color{purple}{\int_0^1 (x(1-x))^{n-1}dx} \\ & = -\frac12\int_0^1 \frac{1}{x(1-x)} \color{blue}{\left(-\sum_{n=1}^\infty \frac{(x(1-x))^{n}}{n}\right)}dx = - \frac12 \int_0^1 \frac{\color{blue}{\ln(1-x(1-x))}}{x(1-x)}dx \\ & = -\frac12 \bigg(\int_0^1 \frac{\ln(1-x(1-x))}{x}dx+\underbrace{\int_0^1 \frac{\ln(1-x(1-x))}{1-x}dx}_{1-x\ \rightarrow \ x}\bigg) \\ & = - \frac12\left(\int_0^1 \frac{\ln(1-x(1-x))}{x}dx +\int_0^1 \frac{\ln(1-(1-x)x)}{x}dx\right) \\ & = - \int_0^1 \frac{\ln(1-x+x^2)}{x}dx = - \int_0^1 \frac{\ln\left(\frac{1+x^3}{1+x}\right)}{x}dx \\ & = \int_0^1 \frac{\ln(1+x)}{x}dx-\underbrace{\int_0^1 \frac{\ln(1+x^3)}{x}dx}_{x^3 \rightarrow x} \\ & = \int_0^1 \frac{\ln(1+x)}{x}dx -\frac13 \int_0^1 \frac{\ln(1+x)}{x}dx =\frac23\int_0^1 \frac{\ln(1+x)}{x}dx \end{align*} $\quad \quad \quad \quad \quad \quad \displaystyle{ =\frac13 \int_0^1 \frac{\ln x}{x-1}dx}$$\displaystyle{=-\frac13\sum_{n=0}^\infty \int_0^1 x^n \ln xdx= \frac13 \sum_{n=0}^\infty \frac{1}{(n+1)^2}=\frac13 \sum_{n=1}^\infty \frac{1}{n^2}}$

As an alternative just take $x=1$ in the following relation shown by Felix Marin: $$ \sum_{n = 1}^{\infty}{x^{n} \over n^{2}{2n \choose n}} =-\int_{0}^{1} \frac{\ln(1-(1-t)tx)}{t} dt. $$

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