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!

Telescoping series

Tags:
  1. Aug 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the formula of the partial sum of the series Ʃ1/[k(k+2)] k from 1 to infinity


    2. Relevant equations



    3. The attempt at a solution

    Using partial fractions i rewrite the series as 1/2*Ʃ[1/k] - [1/(k+2)]

    Then I start writing out the series from k=1 to 5.
    1/2*[(1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+(1/5-1/7)+...+(1/n-1/(n+2))]

    Everything cancels except for 1-1/6-1/7+(1/n-1/(n+2))

    I think that Sn=1/2*(1+1/2-1/(n+2)-1/(n+2))

    My professor in class solved it as 1/2*(1+1/2-1/(n+1)-1/(n+2))

    I don't understand where the 1/n+1 term comes from. Is he right or am I, and why?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 1, 2013 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Apparently that is ##\frac 1 n - \frac 1 {n+2}##. A little spacing would be good.

    Your professor is correct. The last few terms are$$
    ...(\frac 1 {n-3}-\frac 1 {n-1})+(\frac 1 {n-2}-\frac 1 n ) + (\frac 1 {n-1} - \frac 1 {n+1}) +(\frac 1 n -\frac 1 {n+2})$$Only the second term of each of the last two parentheses remains.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Telescoping series
  1. Telescoping Series (Replies: 7)

  2. Telescoping Series (Replies: 1)

Loading...