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.
Best Answer
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.