Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 sis Aug 18 '12 at 7:21
    
@Joseph Skelton: yeah, Raabe's test is also useful here. –  Chris's sis Aug 18 '12 at 7:27
    
It seems that Gauss test may also work nice. –  Chris's sis Aug 18 '12 at 8:01
show 1 more comment

3 Answers

up vote 35 down vote accepted

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.

share|improve this answer
    
really awesome! (+1) –  Chris's sis 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 sis 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  
+1 Painfully simple and elegant. –  DonAntonio Aug 17 '12 at 23:27
add comment

$$ \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}=\frac{(2n)!}{2^{2n}n!^2}\tag{1} $$ 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):

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

share|improve this answer
    
nice! (+1) Thank you. –  Chris's sis Aug 17 '12 at 21:38
    
I'm confused, how do you get (1)? –  Joseph Skelton Aug 18 '12 at 0:53
3  
$$\eqalign{ \frac{{\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!}} = \frac{{\left( {2n} \right)!!\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!\left( {2n} \right)!!}} &\cr = \frac{{\left( {2n} \right)!}}{{\left( {{2^n}n!} \right)\left( {{2^n}n!} \right)}} &\cr = \frac{{\left( {2n} \right)!}}{{{4^n}n{!^2}}} &\cr} $$ –  Pedro 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
add comment

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.

share|improve this answer
    
you mean $4^n$ on the 4th row, right? –  Chris's sis 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 sis Aug 18 '12 at 8:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.