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 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 $\dfrac{\pi^2}6$. Preferably these should be in the spirit of Markus Scheuer's or Jack D'Aurizio's approaches rather than the one similiar by pprime.

Thanks in advance!


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


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!

  • $\begingroup$ 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

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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