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!


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 method 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. $\endgroup$ Commented Dec 24, 2018 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
    Commented Dec 24, 2018 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). $\endgroup$ Commented Dec 24, 2018 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
    Commented Dec 24, 2018 at 13:10
  • $\begingroup$ @braaterAfrikaaner Hey! Here we're again. I still have no problem with you correcting my grammar and spelling but please refrain from changing the tone of my question. I.e. removing my "Thanks in advance!" or phrases like "Quite impressive". While I know that these aren't really necessary I like to include them anyways as they represent the way I interact with other people in real life. $\endgroup$
    – mrtaurho
    Commented Sep 5, 2021 at 20:44

1 Answer 1


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: $$ \sum_{n = 1}^{\infty}{x^{n} \over n^{2}{2n \choose n}} =-\int_{0}^{1} \frac{\ln(1-(1-t)tx)}{t} dt. $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .