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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.

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    Demonstrate with logs --> Taylor?2012-08-18

3 Answers 3

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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.

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    The estimate is simple and loose, but it gets the job done (+1).2012-08-17
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$ \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} $

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    @PeterTamaroff: Thanks for that. I have just added a derivation to the answer.2012-08-18
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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.

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    OK. Thank you for your nice solution! (+1)2012-08-18