The series$\sum\limits_{n=1}^{\infty}(\frac{(2n-1)!!}{(2n)!!})^p$ converges if and only if $p>2$

Question

Problem: The series$\sum\limits_{n=1}^{\infty}(\frac{(2n-1)!!}{(2n)!!})^p$ converges if and only if $p>2$

My question is: can anyone solve the problem by using binomial coefficients or Stirling's formula?

Answer 1

Observe that $$ \frac{(2n-1)!!}{(2n)!!} = \frac{(2n-1)!! (2n)!! }{(2n)!! (2n)!!} = \frac{(2n)! }{ 2^{2n} (n!)^2 } \sim (\text{ using Stirling's formula}) \\ \frac{\sqrt{2\pi 2n} (2n)^{2n} e^{-2n} }{ 2^{2n} (2\pi n) n^{2n} e^{-2n} }= \frac{1}{\sqrt{\pi n}}. $$ Hence, you effectively have a series of $(\frac{1}{n^{1/2}})^p$ and need $p/2 >1$ for convergence. You may use comparison tests to make things rigorous (namely, the passage from the original term to reduction by Stirling).