# 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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