# Proof that ratio of the products of odd and even numbers converges

1. Sep 23, 2012

### Rasmus

1. The problem statement, all variables and given/known data
Show that

$a_{n} = \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n - 1)}{2 \cdot 4 \cdot 6 \cdot ... \cdot 2n}$

converges when n $\rightarrow$∞

and n is a natural number

2. Relevant equations
None that I can think of.

3. The attempt at a solution

This was from an exam and I was pretty much stumped.

Last edited: Sep 23, 2012
2. Sep 23, 2012

### jbunniii

What can you say about $a_{n+1}$ vs. $a_{n}$? Which one is larger?

3. Sep 23, 2012

### Rasmus

Thanks for the hint!

As far as I understand the problem we start by 1/2 and multiply another rational for each increment increase in n. Meaning that $a_{n+1}$ = $a_{n}$ * (2n)/(2n +1). Since n is a natural number that menas the latter factor is less than one, therefore $a_{n}$ must be greater than $a_{n+1}$

I'm pretty sure I can work with this, but right now I need to be away from the computer for an hour or so.

So far I'm thinking maybe proof by induction to show that each n must be smaller than the next and to use the fact that they're natural numbers to put a lower limit of zero on the product. And then squeeze $a_{n}$ to zero.

Thanks a lot

4. Sep 23, 2012

### Ray Vickson

The denominator is $2 \cdot 4 \cdot 6 \cdots 2n = 2^n n!$ and the numerator is
$$1 \cdot 3 \cdot 5 \cdots (2n-1) = \frac{(2n-1)!}{2 \cdot 4 \cdots (2n-2)} = \frac{(2n-1)!}{2^{n-1} (n-1)!}.$$ Thus,
$$a_n = \frac{(2n-1)!}{2^{n-1} (n-1)!} \cdot \frac{1}{2^n n!}.$$ Now use Stirling's formula.

RGV

5. Sep 23, 2012

### jbunniii

You don't need to "squeeze $a_n$ to zero." All the problem asks you to do is to show that the sequence converges. This is easier than finding what value it converges to.

You established that $a_n$ is a monotonically decreasing sequence. What do you know about convergence of monotone sequences?

Last edited: Sep 23, 2012