$$\sum\limits_{n\ge1}\frac1{n^2}=\sum\limits_{n\ge1}\frac3{n^2\binom{2n}n}\tag1$$
Note that $(1)$ holds since the LHS is given by $\zeta(2)$ whereas the RHS by $6\arcsin^2 1$ which both equal $\dfrac{\pi^2}6$ as it is well-known. However, I am interested in proving $(1)$ 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 attempts. For example within Jack D'Aurizio's notes there is a way proposed exploiting creative telescoping $($see page $5$f.$)$, which I do not understand completely (yet).
Hence I have come across a proof of a similar 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 $(1)$ besides the ones mentioned which do not rely on actually showing that they both equal $\dfrac{\pi^2}6$. Preferably these should be in the spirit of Markus Scheuer's or Jack D'Aurizio's approaches rather than the one similar by pprime.
Thanks in advance!
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 method can be found which makes it three possibilities provided by Jack alone. Quite impressive!
Best Answer
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. $$