Convergence\Divergence of $\sum\limits_{n=1}^{\infty}\frac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)}$

Since $$ \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}{2 \cdot 4 \cdot \ldots \cdot (2n)} \ge \frac{1 \cdot 2 \cdot 4 \cdot \ldots \cdot (2n-2)}{2 \cdot 4 \cdot \ldots \cdot (2n)} = \frac1{2n} $$ the series diverges by comparison to the Harmonic series.

$$ \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}=\frac{(2n)!}{2^{2n}n!^2}\tag{1} $$ Using Stirling's Formula, we get that $$ \frac{(2n)!}{2^{2n}n!^2}\sim\frac1{\sqrt{\pi n}}\tag{2} $$ By the $p$-test, $$ \sum_{n=1}^\infty \frac1{n^p}\tag{3} $$ diverges for $p\le1$, $$ \sum_{n=1}^\infty\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\tag{4} $$ diverges.

Derivation of (1):

$$ \begin{align} \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)} &=\frac{1\cdot\color{#C00000}{2}\cdot3\cdot\color{#C00000}{4}\cdot5\cdot\color{#C00000}{6}\cdots(2n-1)\cdot\color{#C00000}{(2n)}}{2\cdot4\cdot6\cdots(2n)\color{#C00000}{2\cdot4\cdot6\cdots(2n)}}\\ &=\frac{(2n)!}{(2^nn!)^2} \end{align} $$