2009/07/29 · 18:54
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whole sum of the reciprocal Catalan numbers


In this little post I complete the details of the calculation

2+\frac{4\sqrt{3}\pi}{27}=\sum_{n=0}^{\infty}\frac{n+1}{{2n\choose n}}=1+1+\frac{1}{2}+\frac{1}{5}+\frac{1}{14}+\frac{1}{42}+\cdots

for the reciprocals C_n=\frac{1}{n+1}{2n\choose n}, the famous so called Catalan numbers. It seems this is well known but it is scarcely quoted anywhere: Cf1, Cf2

We begin by recalling the relation {C_n}^{-1}=(2n+1)(n+1)\int_0^1t^n(1-t)^n{\rm{d}}t, that is,  in terms of the Beta function. So

\sum_{n=0}^{\infty}{C_n}^{-1}=\sum_{n=0}^{\infty}(2n+1)(n+1)\int_0^1t^n(1-t)^n{\rm{d}}t

=\int_0^1\left[\sum_{n=0}^{\infty}(2n+1)(n+1)t^n(1-t)^n\right]{\rm{d}}t

=\int_0^1\frac{1+3t - 3t^2}{{(1-t+t^2)}^3}{\rm{d}}t

=2+\frac{4\sqrt{3}\pi}{27}

This was hinted to me, thanks to, by the professor Qiaochu Yuan and I am not ashamed to confess that the calculations were executed in mathematica-v5.  Why? well, the risk of introducing errors is a little-big.

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One response to “whole sum of the reciprocal Catalan numbers”

  1. Anonymous
    2019/10/08 at 02:50

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