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

  1. Mar 18, 2005 #1
    I have no idea how to do this :'( really the only part I don't understand is the ending part.....like for

    the infinite sum of (1/n(n+1)).............I know you start of by partial fractions.....then you just plug in a few numbers for n.

    so I end up with:

    (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1))

    then in another problem...the infinite sum of ( 2/{(n-1)(n+1)} ) ends up like...

    (1 - 1/3) + (1/2 - 1/4) +1/3 - 1/5) + ... + (1/(n-3) - 1/(n-1)) + (1/(n-2) - 1/n)


    I want to know how you get the end results.....the ones with "n" in them....I don't understand how to get those numbers....or why they are what they are....

    I understand the first part without the "n"....but I don't know how to end it with the "n"....I hope this makes since. :confused:
     
  2. jcsd
  3. Mar 18, 2005 #2

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    OK, let's take a look at:
    [tex]\sum_{n=1}^{4}\frac{1}{n}-\frac{1}{n+1}[/tex]
    So if we write it out longhand it turns into:
    [tex](\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})[/tex]
    but we can regroup:
    [tex]\frac{1}{1}+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+(-\frac{1}{4}+\frac{1}{4})-\frac{1}{5}[/tex]
    This is just moving the parens around, but the paired numbers nicely all zero out, leaving you with
    [tex]\frac{1}{1}-\frac{1}{5}[/tex]

    It should be easy to see that the pattern can be extended to any number of terms so that you end up with
    [tex]\sum_{n=1}^{k}\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{k+1}[/tex]

    There are more formal approaches to this, but I doubt that they will be particularly helpful to you.
     
    Last edited: Mar 18, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Telescoping Series
  1. Fourier Series (Replies: 1)

  2. Telescoping sum (Replies: 4)

  3. Taylor series (Replies: 5)

Loading...