# 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.

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 Aryabhata's estimate $\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)} \ge \frac{1}{\sqrt{4n}}$ gives the answer by the $p$-test (as in my answer below). – robjohn♦ Aug 17 '12 at 21:43 You can also use Raabe's test, $lim_{n \to \infty} n\left(\frac{a_n}{a_{n+1}}-1\right)$ if your result $P>1$ then the series is convergent, $P<1$ the series is divergent, and if $P=1$ then the test is inconclusive. In solving the limit, you are almost assuredly going to have to use L'Hospital rule mathworld.wolfram.com/RaabesTest.html – Joseph Skelton Aug 18 '12 at 0:59 @robjohn: when I firstly saw that inequality I wondered where it came from. I don't remember exactly, but it's possible that I met it before, maybe in my first year in high school when studying the induction chapter. This inequality form is really nice. – Chris's wise sister Aug 18 '12 at 7:21 @Joseph Skelton: yeah, Raabe's test is also useful here. – Chris's wise sister Aug 18 '12 at 7:27 It seems that Gauss test may also work nice. – Chris's wise sister Aug 18 '12 at 8:01
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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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really awesome! (+1) – Chris's wise sister Aug 17 '12 at 21:23
You beat me! Good job. – Byron Schmuland Aug 17 '12 at 21:24
@Morgan Sherman: it's nice to see that such a small trick brings you to the result so fast. :-) – Chris's wise sister Aug 17 '12 at 21:31
The estimate is simple and loose, but it gets the job done (+1). – robjohn Aug 17 '12 at 21:45
+1 Painfully simple and elegant. – DonAntonio Aug 17 '12 at 23:27

$$\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}=\frac{(2n)!}{2^{2n}n!^2}\tag{1}$$ 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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nice! (+1) Thank you. – Chris's wise sister Aug 17 '12 at 21:38
I'm confused, how do you get (1)? – Joseph Skelton Aug 18 '12 at 0:53
\eqalign{ \frac{{\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!}} = \frac{{\left( {2n} \right)!!\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!\left( {2n} \right)!!}} &\cr = \frac{{\left( {2n} \right)!}}{{\left( {{2^n}n!} \right)\left( {{2^n}n!} \right)}} &\cr = \frac{{\left( {2n} \right)!}}{{{4^n}n{!^2}}} &\cr} – Peter Tamaroff Aug 18 '12 at 1:48
ah, ok. That makes much more sense now. I was unaware there was even a name for that product. Thanks! – Joseph Skelton Aug 18 '12 at 2:30
@PeterTamaroff: Thanks for that. I have just added a derivation to the answer. – robjohn Aug 18 '12 at 7:32

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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 you mean $4^n$ on the 4th row, right? – Chris's wise sister Aug 18 '12 at 8:18 Yeah, edited. Thanks. – Rijul Saini Aug 18 '12 at 8:23 OK. Thank you for your nice solution! (+1) – Chris's wise sister Aug 18 '12 at 8:23