Convergence\Divergence of $\sum\limits_{n=1}^{\infty}\frac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)}$ Prove convergence\divergence of the series:
$$\sum_{n=1}^{\infty}\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)}$$
Here is what I have at the moment:
Method I
My first way uses a result that is related to Wallis product that we'll denote by $W_{n}$.  Also,
we  may denote $\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)}$ by $P_{n}$. Having noted these and taking a large
value of $n$
we get:
$$(P_{n})^2 =\frac{1}{W_{n} \cdot (2n+1)}\approx\frac{2}{\pi}\cdot \frac{1}{2n+1}$$
$$P_{n}\approx \sqrt {\frac{2}{\pi}} \cdot \frac{1}{\sqrt{2n+1}}$$ 
Further we have that:
$$\lim_{n\to\infty}\sqrt {\frac{2}{\pi}} \cdot \frac{n}{\sqrt{2n+1}} \le \sum_{n=1}^{\infty} P_{n}$$ 
that obviously shows us that the series diverges.
Method II
The second way is to resort to the powerful Kummer's Test and firstly proceed with the ratio test:
$$\lim_{n\to\infty} \frac{P_{n+1}}{P_{n}}=\frac{2n+1}{2n+2}=1$$
and according to the result, the ratio test is inconclusive.
Now, we apply Kummer's test and get:
$$\lim_{n\to\infty} \frac{P_{n}}{P_{n+1}}n-(n+1)=\lim_{n\to\infty} -\frac{n+1}{2n+1}=-\frac{1}{2} \le 0$$
Since 
$$\sum_{n=1}^{\infty} \frac{1}{n} \longrightarrow \infty$$
our series diverges and we're done.
On the site I've also found  a related question with answers that can be applied for my question. 
Since I've already have some answers for my question you may regard it as a recreational one and if you have a nice proof to share I'd be glad to receive it. I like this question very much and want to make up a collection with nice proofs for it. Thanks. 
 A: $$
\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}=\frac{(2n)!}{2^{2n}n!^2}\tag{1}
$$
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):
$$
\begin{align}
\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}
&=\frac{1\cdot\color{#C00000}{2}\cdot3\cdot\color{#C00000}{4}\cdot5\cdot\color{#C00000}{6}\cdots(2n-1)\cdot\color{#C00000}{(2n)}}{2\cdot4\cdot6\cdots(2n)\color{#C00000}{2\cdot4\cdot6\cdots(2n)}}\\
&=\frac{(2n)!}{(2^nn!)^2}
\end{align}
$$
A: 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.
A: As robjohn notes, $$
\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}=\frac{(2n)!}{2^{2n}n!^2} = \frac 1{4^n} \binom{2n}{n}
$$
Noting that $$(2n+1) \binom{2n}{n} > \sum_{i=0}^{2n} \binom{2n}{i} = 4^n$$
As $\binom{2n}{n}$ is the largest binomial coefficient.
Therefore, $$\frac 1{4^n} \binom{2n}{n} > \frac{1}{2n+1},$$ and hence the series diverges, by the comparison test.
