4
$\begingroup$

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.

$\endgroup$
5
  • 6
    $\begingroup$ Mathematica knows both. $\endgroup$ Aug 13, 2019 at 17:08
  • 8
    $\begingroup$ 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
  • 3
    $\begingroup$ 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
  • 3
    $\begingroup$ 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

2
$\begingroup$

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 $0 \leq x \leq 4$

$\endgroup$
1
  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.