<s>[INST] Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $ 
How to find infinite sum How to find infinite sum $$1+\dfrac13+\dfrac{1\cdot3}{3\cdot6}+\dfrac{1\cdot3\cdot5}{3\cdot6\cdot9}+\dfrac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $$

I can see that 3 cancels out after 1/3, but what next? I can't go further.
 [/INST] From the OGF of Catalan numbers we have that:
$$ \sum_{n\geq 0}\binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}} $$
where the radius of convergence of the LHS is $\frac{1}{4}$ since $\frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}=\frac{4n+2}{n+1}$.
By evaluating the previous identity at $x=\frac{1}{6}$ it follows that:
$$ \color{red}{\sqrt{3}} = 1+\sum_{n\geq 1}\frac{(2n)!}{n! n! 6^n} = 1+\sum_{n\geq 1}\frac{(2n-1)!!}{3^n n!} $$
where the RHS is exactly our sum.
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