Convergence of the next series I'm trying to determine the convergence of this series:
$$\sum \limits_{n=1}^\infty\left(\frac12·\frac34·\frac56·...\frac{2n-3}{2n-2}·\frac{2n-1}{2n}\right)^a$$
I've tried using D'Alambert criteria for solving it.
$$\lim_{n->\infty}\frac{(\frac12·\frac34·\frac56·...\frac{2n-3}{2n-2}·\frac{2n-1}{2n}\frac{2n}{2n+1})^a}{(\frac12·\frac34·\frac56·...\frac{2n-3}{2n-2}·\frac{2n-1}{2n})^a} =
 \lim_{n->\infty}\left(\frac{(\frac12·\frac34·\frac56·...\frac{2n-3}{2n-2}·\frac{2n-1}{2n}·\frac{2n}{2n+1})}{(\frac12·\frac34·\frac56·...\frac{2n-3}{2n-2}·\frac{2n-1}{2n})}\right)^a$$
Which becomes:
$$\lim_{n->\infty}\left(\frac{2n}{2n+1}\right)^a$$
But after that, the limit is 1, so its convergence is unknown.
Any idea?

Rewrite the summand in terms of factorials like this: $$ \left( \frac{(2n)!}{ 2^{2n} (n!)^2} \right)^a .$$ Applying Stirling's approximation gives $$ \frac{(2n)!}{ 2^{2n} (n!)^2}  \sim \frac{1}{\sqrt{\pi n} } $$ so to finish off, apply what you know about the convergence of $ \displaystyle \sum \frac{1}{n^p} $ for various $p.$
