Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Compute the sum

  1. Oct 14, 2004 #1
    Compute the following sum
    when n goes from 1 to oo of
    1/(n^2(2n-1))


    So far this is what i've done
    1/(n^2(2n-1))=4/(2n(2n-1))-1/n^2
    I know the sum from 1 to oo of 1/n^2 =(pi^2)/6
    So I get
    1/(n^2(2n-1))=4/(2n(2n-1)) - pi^2/6

    so I need to compute the sum of 4/(2n(2n-1)) which i have no idea how to do.

    Any hints or tips would be great
     
  2. jcsd
  3. Oct 14, 2004 #2
    Assuming u did the first part correctly,
    for ur question,
    use partial fractions,
    4/((2n)(2n-1))
    = (4/(2n-1))-(4/(2n))

    Now write the first few numbers of the series in this fashion and see if can notice something peculiar ...

    -- AI
     
  4. Oct 14, 2004 #3
    thankyou the reason I couldn't do the question is because I went
    4/(2n(2n-1))=4/(2n-1)-2/n

    and I couldn't see any apparent pattern.

    I will try your partial fraction right now. Thankyou
     
  5. Oct 15, 2004 #4
    by teh way i was wondering if i got your question right is it 1 / n ^(4n-2)

    or is it 1/n^2 * (2n-1)

    in any case the first one does converge since n^2 / n^4n < n^2 / n^4 = 1 / n^2 which is a convergent p series
    the second one you can use the limit comparison test to show that it is lesser the 1/n^2 which is a convergent p series
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Compute the sum
  1. An infinite sum (Replies: 3)

  2. Telescoping sum (Replies: 4)

Loading...