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While solving for that first an excercise is to find the proof for $\zeta(2)=3\sum_{n=1}^\infty\frac{1}{n^2\binom{2n}{n}}$.

Here my first thought being, $\zeta(2)=\sum_{n=1}^\infty\frac{1}{n^2}$ which we are asked to equate with the above expression, so does that mean inside the summation somehow vaguely $3$ cancels out with the terms $\frac{1}{\binom{2n}{n}}$.

Now I think if my calculations were correct the expansion terms also doesnt really match up. When I also tried to find this how to do this and stuff I believe the expression, $3\sum_{n=1}^\infty\frac{1}{n^2\binom{2n}{n}}$ actually called or be expressed as Diriclet Beta function where $\beta(2)$ gives this expression.

Now I am totally confused whats happening, I believe there exist some very close relation between zeta expression and beta expression using the Bernoulis number and Eulers numbers consequently but, that again just fuzzes my brain. What am I missing.

Btw, there is the mention of using $(-1)^{n-1}$ instead of dividing by $n$, I am not able to substitute.

Arnab Tapadar
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  • More generally, see this post. – Dietrich Burde Jun 27 '25 at 17:43
  • Actually I am having difficulty following the Browein Bradly paper which was mentioned and I believe they tend to use the Beuker like integrals as for the generalised like $n^{2k}\times \binom{2n}{n}$. Actually I am asked by my professor to use the manipulations as Apery used for $\zeta(3)$. Thanks for the reccomended post @DietrichBurde – Arnab Tapadar Jun 28 '25 at 14:31

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