1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Sep 23, 2012 #1
    1. The problem statement, all variables and given/known data
    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

    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. jcsd
  3. Sep 23, 2012 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What can you say about [itex]a_{n+1}[/itex] vs. [itex]a_{n}[/itex]? Which one is larger?
     
  4. Sep 23, 2012 #3
    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
     
  5. Sep 23, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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
     
  6. Sep 23, 2012 #5

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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?
     
    Last edited: Sep 23, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof that ratio of the products of odd and even numbers converges
  1. Even/odd case proof (Replies: 1)

Loading...