1. Not finding help here? Sign up for a free 30min 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!

Determine the convergence or divergence of the sequence

  1. Jul 31, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine the convergence or divergence of the sequence with the given nth term. If the sequence converges find its limit.

    an = (1*3*5*....*(2n-1))/(2n)n


    2. Relevant equations
    lim n->infinity an = L


    3. The attempt at a solution
    The answer in the book shows:
    1/2n * 3/2n * 5/2n....(2n-1)/2n less than 1/2n

    So, lim n->infinity an = 0, converges

    I don't understand where the 1/2n * 3/2n * 5/2n....(2n-1)/2n less than 1/2n comes from. I'm obviously missing something. Could anyone please explain that part to me? I'd really appreciate any help at all. Thanks :)
     
  2. jcsd
  3. Jul 31, 2013 #2

    hilbert2

    User Avatar
    Science Advisor
    Gold Member

    Obviously ##1 \cdot 2 \cdot 3 \dots (2n-1) \cdot (2n) < (2n)^{n}##, because in the first product all factors except the last one are smaller than in the second product.

    Therefore, also ##\frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{(2n)^{n}}<\frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{1 \cdot 2 \cdot 3 \dots (2n-1) \cdot (2n)}=\frac{1}{2 \cdot 4 \dots (2n)}##.

    Can you now tell that the sequence is convergent?
     
  4. Jul 31, 2013 #3

    pasmith

    User Avatar
    Homework Helper

    For n sufficiently large, 0 < 1/2n < 3/2n < 5/2n < ... < (2n-1)/2n < 1. So you're multiplying 1/(2n) by something which is strictly less than 1, which means that the result is strictly less than 1/(2n).
     
  5. Jul 31, 2013 #4
    Thanks for your help, but I'm still not getting it. When I see the sequence an = (1*3*5*....*(2n-1))/(2n)^n I would think it looks something like 1/2(1)^1 * 3/2(2)^2 * 5/2(3)^3 .... Is that wrong? I don't understand where the exponent n is going from the denominator.
     
  6. Jul 31, 2013 #5

    pasmith

    User Avatar
    Homework Helper

    Yes; it's
    [tex]
    \frac1{2n} \times \frac3{2n} \times \cdots \times \frac{2n-1}{2n} = \frac{1 \times 3 \times \cdots \times (2n-1)}{(2n)^n}
    [/tex]
    since there are [itex]n[/itex] factors and the denominator is the same for each.
     
  7. Jul 31, 2013 #6

    Mark44

    Staff: Mentor

    Both, if you mean ##\frac 1 {2n}##, ##\frac 3 {2n}##, and so on, you need parentheses around each denominator. Otherwise what both of you wrote would be interpreted as (1/2) * n, (3/2) * n, etc.
     
    Last edited: Jul 31, 2013
  8. Jul 31, 2013 #7
    Ok now I understand why those two equations are equal, thanks. It seems so obvious now. What I still don't understand is how does saying 1/2n * 3/2n * 5/2n....(2n-1)/2n less than 1/2n help me conclude that the sequence diverges?
     
  9. Jul 31, 2013 #8

    hilbert2

    User Avatar
    Science Advisor
    Gold Member

    ^ Actually, the sequence converges, because you can make the quotient arbitrarily close to zero by choosing sufficiently large n.
     
  10. Jul 31, 2013 #9

    pasmith

    User Avatar
    Homework Helper

    If [itex]\epsilon > 0[/itex], does there exist a positive integer [itex]N[/itex] such that [itex]1/(2N) < \epsilon[/itex]?

    What does that tell you about the convergence or divergence of the sequence?
     
  11. Jul 31, 2013 #10
    Oh ok so the limit as n goes to infinity is zero which means it converges, but I'm a little lost. How do I know the limit is zero? Thanks for all your help I really appreciate it :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted