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

Rasmus
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Homework Statement


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

Homework Equations


None that I can think of.

The Attempt at a Solution



This was from an exam and I was pretty much stumped.
 
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What can you say about a_{n+1} vs. a_{n}? Which one is larger?
 
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
 
Rasmus said:

Homework Statement


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

Homework Equations


None that I can think of.


The Attempt at a Solution



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

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)}<br /> = \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
 
Rasmus said:
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.
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?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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