Infinite sum power series

Question

I would like to show $$ \sum_{r=0}^{\infty}\frac{1}{6^r} \binom{2r}{r}= \sqrt{3} $$ I have tried proving this using telescoping sum, limit of a sum, and some combinatorial properties but I couldn't solve it.

Answer 1

Welcome to MSE.

I'll prove a more general statement: for $|M|<1$, $$ \sum _{r=0}^{\infty} \binom{2r}{r} (M/4)^r = (1-M)^{-1/2} $$There are some really nice proofs using generating functions, but mine will be somewhat direct (read: coarse). First, note that using the half-binomial identities, we can rewrite the summand as $$ \sum _{r=0}^{\infty} \binom{2r}{r} (M/4)^r = \sum _{r=0}^{\infty} \binom{-1/2}{r} (-M)^r $$Then the result follows from the generalized binomial theorem. $$ \sum _{r=0}^{\infty} \binom{\color{red}{-1/2}}{r} (\color{blue}{-M})^r = (1+(\color{blue}{-M}))^{\color{red}{-1/2}} $$In particular, $$ \sum _{r=0}^{\infty} \binom{2r}{r}( (\color{green}{2/3})/4)^r = (1-\color{green}{2/3})^{-1/2}=\sqrt{3} $$

Answer 2

Hint: $\sum_{r\ge 0} \binom{2r}{r} z^r = \frac{1}{\sqrt{1-4z}}$ for $|z|<1/4$.