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

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Homework Help Overview

The discussion revolves around the convergence of the sequence defined by the ratio of the products of odd and even natural numbers, specifically \( a_{n} = \frac{1 \cdot 3 \cdot 5 \cdots (2n - 1)}{2 \cdot 4 \cdot 6 \cdots 2n} \) as \( n \) approaches infinity. Participants are exploring the properties of this sequence and its behavior as \( n \) increases.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between \( a_{n+1} \) and \( a_{n} \), questioning which is larger and how this affects convergence. There are considerations of using proof by induction and the application of Stirling's formula to analyze the sequence further.

Discussion Status

The discussion is active, with participants providing hints and exploring various approaches to demonstrate convergence. Some have noted the sequence is monotonically decreasing, which may relate to its convergence properties. There is no explicit consensus on a final method or conclusion yet.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on understanding the nature of the sequence rather than calculating its limit.

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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