calculus - 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?

Answer

Let $$a_n={1\over2}\cdot{3\over4}\cdot{5\over6}\cdot\ \cdots\ \cdot {{2n-3}\over{2n-2}}\cdot{{2n-1}\over{2n}}.$$

Note that $a_{n+1}=a_n\cdot {2n +1\over2(n+1)}$.

We will show that for all $n$:
$$
{1\over\sqrt{ 4n}} \le a_n\le {1\over\sqrt{ 2n+1}}.
$$
Having established this, it will follow by comparison with $p$-series that
$\sum\limits_{n=1}^\infty a_n^a$ converges if and only if $a>2$.

We establish the inequalities by induction.

Towards showing that $ a_n\le {1\over\sqrt{ 2n+1}}$, first

note that $a_1\le {1\over \sqrt{2\cdot 1+1}}$.

Now
suppose $a_n \le {1\over\sqrt{ 2n+1}}$.
Then $$ a_{n+1}^2\le \biggl[\,{2n +1\over 2(n+1)}\biggr]^2\cdot{1\over 2n+1}
={2n+1\over 4(n+1)^2}={2n+1\over 4n^2+8n+4}\le{2n+1\over 4n^2+8n+1 }={1\over 2n+1}.$$

So, $a_n\le {1\over\sqrt {2n+1}}$ for all $n$.

Towards showing that $ {1\over\sqrt{ 4n}} \le a_n$, first

note that $a_1\ge {1\over \sqrt{4\cdot 1 }}$.

Now
suppose $a_n \ge {1\over\sqrt{ 4n}}$.
Then $a_{n+1}^2\ge \Bigl[\,{2n +1\over 2(n+1)}\,\Bigr]^2\cdot{1\over 4n}$.
Now
$$
\biggl[\,{2n +1\over 2(n+1)}\,\biggr]^2\cdot{1\over 4n}-{1\over 4(n+1)}
={1\over 16 n(n+1)^2}>0;$$
thus

$$
a_{n+1}^2\ge{1\over 4(n+1)}.
$$

So, $a_n\ge {1\over\sqrt {4n}}$ for all $n$.