Let $\zeta(n)$ denote the Riemann zeta function defined for positive integers greater than $1$ by its usual infinite series. Thus, $\zeta(2)=\sum_{k=1}^\infty\frac{1}{k^2}$. Many formulae exist involving $\zeta(2)$, including the Apéry-like fast-converging series: $$ \zeta (2)=3\sum _{{n=1}}^{{\infty }}{\frac {1}{n^{{2}}{\binom {2n}{n}}}}. $$ Recently I have found the following similar-looking series:

$$ \zeta (2)=\frac83\sum _{{n=1}}^{{\infty }}{\frac {2^{n-1}}{n^{{2}}{\binom {2n}{n}}}}, $$ $$ \zeta (2)=\frac94\sum _{{n=1}}^{{\infty }}{\frac {3^{n-1}}{n^{{2}}{\binom {2n}{n}}}} $$ and $$ \zeta (2)=\frac43\sum _{{n=1}}^{{\infty }}{\frac {4^{n-1}}{n^{{2}}{\binom {2n}{n}}}}. $$

Are these series already known? A quick internet search yields no such results.

EDIT forgot to add the second series.

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    $\begingroup$ Mathematica knows both. $\endgroup$ Aug 13, 2019 at 17:08
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    $\begingroup$ We have: $$\sum_{n=1}^\infty \frac{x^n}{n^2 \binom{2n}{n}}=2\arcsin^2 \frac{\sqrt{x}}{2}$$ You can check your particular cases using this general formula $\endgroup$
    – Yuriy S
    Aug 16, 2019 at 11:35
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    $\begingroup$ To answer the question, no, these results are not new, since the general Taylor series for $\arcsin^2 y$ has been known for a long time $\endgroup$
    – Yuriy S
    Aug 16, 2019 at 11:38
  • $\begingroup$ @YuriyS Thank you for the comment! If you could elaborate on that in the form of an answer I will be happy to accept it $\endgroup$
    – Klangen
    Aug 16, 2019 at 11:42
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    $\begingroup$ As a (rather silly) instance of this formula, one may write $$\zeta{2} = \frac{25}{12}\sum_{n=1}^\infty \frac{1}{n^2 \binom{2n}{n}} \left(\frac{{5}-\sqrt{5}}{2}\right)^n.$$ $\endgroup$ Aug 16, 2019 at 20:49

1 Answer 1


Please see the following answer for the proof of the fact

$\displaystyle \sum_{n=1}^\infty \frac{x^n}{n^2 \binom{2n}{n}}=2\left[\sin^{-1} \frac{\sqrt{x}}{2} \right]^2$

Just note that in the result given in answer, you need to replace $x$ by $\frac{\sqrt{x}}{2}$ and also make sure that $0 \leq x \leq 4$

  • $\begingroup$ @PTDS can you please elaborate in (much) more detail? I cannot offer the bounty to a "hint"... $\endgroup$
    – Klangen
    Aug 24, 2019 at 8:31

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