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!

Convergence/Divergence of series

  1. Nov 25, 2007 #1

    I have to prove conv/div. for the following series:


    I use the "ratio-test" and get the following:

    [tex]\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}} = \lim_{n\to\infty} \frac{(2n+2)!}{(2n)!} \frac{n^n}{(n+1)^{n+1}} = \lim_{n\to\infty} \frac{(2n+2)(2n+1)}{(n+1)} (\frac{n}{1+n})^n = \infty \frac{1}{e} = \infty[/tex]

    This means the series diverges, however, the series should converge (I could find the finite sum online).

    Where is my mistake?

    Thank you!
  2. jcsd
  3. Nov 26, 2007 #2
    The last half terms can be written as {(n+l)/n}{(n+2)/n}{(n+3)/n}........{2n/n}, and the first n terms are just n!.
  4. Nov 26, 2007 #3
    Sorry, I do not quite understand what you mean or how this helps me. Could you elaborate on that?

    Also, I am interested in why the test I am using does not work out like it should or if I made an algebra mistake somewhere along the way.
  5. Nov 26, 2007 #4

    Gib Z

    User Avatar
    Homework Helper

    Using stirlings approximation to replace the factorial, I get the series diverges. Where did you find online its sum?
  6. Nov 26, 2007 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What he's saying is that if you split it up, you get 1/n*1/n*1/n...*(2n)(2n-1)(2n-2)...(n+1)*n!

    So you put one n under each 2n-k and get


    As each (2n-k)/n>1, and n!>1, each term in the series is >1. So there's very little reason why it would converge
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook