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

  • Thread starter Rasmus
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  • #1
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Homework Statement


Show that

[itex]a_{n} = \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n - 1)}{2 \cdot 4 \cdot 6 \cdot ... \cdot 2n}[/itex]

converges when n [itex]\rightarrow[/itex]∞

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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Answers and Replies

  • #2
jbunniii
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What can you say about [itex]a_{n+1}[/itex] vs. [itex]a_{n}[/itex]? Which one is larger?
 
  • #3
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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 [itex]a_{n+1}[/itex] = [itex]a_{n}[/itex] * (2n)/(2n +1). Since n is a natural number that menas the latter factor is less than one, therefore [itex]a_{n}[/itex] must be greater than [itex]a_{n+1}[/itex]

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 [itex]a_{n}[/itex] to zero.

Thanks a lot
 
  • #4
Ray Vickson
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Homework Statement


Show that

[itex]a_{n} = \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n - 1)}{2 \cdot 4 \cdot 6 \cdot ... \cdot 2n}[/itex]

converges when n [itex]\rightarrow[/itex]∞

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 [itex] 2 \cdot 4 \cdot 6 \cdots 2n = 2^n n! [/itex] and the numerator is
[tex] 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)!}.[/tex] Thus,
[tex] a_n = \frac{(2n-1)!}{2^{n-1} (n-1)!} \cdot \frac{1}{2^n n!}.[/tex] Now use Stirling's formula.

RGV
 
  • #5
jbunniii
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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 [itex]a_{n}[/itex] to zero.
You don't need to "squeeze [itex]a_n[/itex] 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 [itex]a_n[/itex] is a monotonically decreasing sequence. What do you know about convergence of monotone sequences?
 
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