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 for Infinite Series

  1. Dec 28, 2008 #1
    1. Examine the series [tex]\frac{1}{1 . 2} +[/tex][tex]\frac{1}{2 . 3}+[/tex][tex]\frac{1}{3 . 4}+[/tex][tex]\frac{1}{4 . 5}...[/tex] for convergence.

    3. The attempt at a solution

    The following is the book's answer:

    "[tex]lim_{n\rightarrow \infty}S_{n}[/tex]
    [tex]lim_{n\rightarrow \infty} (1 - \frac{1}{n + 1}) = 1 - 0 = 1[/tex]
    Hence the series converge and its sum is 1. "

    From [tex]S_{n} = [/tex][tex]\frac{1}{1 . 2} +[/tex][tex]\frac{1}{2 . 3}+[/tex][tex]\frac{1}{3 . 4}+[/tex][tex]\frac{1}{4 . 5}...[/tex]

    I can see that the nth term is [tex]\frac{1}{n . (n+1)}[/tex] but I can't follow how the book's obtained the "[tex]1 - \frac{1}{n + 1}[/tex]" or th nth partial sum.

    I appreciate some help. Unfortunently there are no explainations in the book on this question.

  2. jcsd
  3. Dec 28, 2008 #2


    User Avatar
    Homework Helper

    try splitting [itex]\frac{1}{n . (n+1)}[/itex] into partial fractions, then use the property of a telescoping series.
  4. Dec 29, 2008 #3
    Is this correct:

    [tex](\frac{1}{1} \times \frac{1}{2}) + (\frac{1}{2} \times \frac{1}{3}) + (\frac{1}{3} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{1}{5}) + ... + (\frac{1}{n} \times \frac{1}{n + 1})[/tex]

    I still don't understand how to reach [tex]1 - \frac{1}{n + 1}[/tex]
  5. Dec 29, 2008 #4
    Did you rewrite 1/[n(n+1)] using partial fractions? Doesn't seem like you have. You just wrote out the first few terms and the n-th term.
  6. Dec 29, 2008 #5

    Now you got:

    [tex](1 - \frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{n}-\frac{1}{n+1})[/tex]

    What can you assume?
  7. Dec 29, 2008 #6
    Oh, thanks. Now I understand what rockfreak was saying about splitting it into partial fractions.

    Now I can test for convergence in [tex]lim_{n \rightarrow \infty}(\frac{1}{n} - \frac{1}{n+a})[/tex] but the book says we need to have "[tex]1 - \frac{1}{n+1}[/tex]" & I'm not sure where I can find it from.
  8. Dec 29, 2008 #7
    No... look at what he wrote you see how for -1/2 there's a 1/2, for -1/3 there's a 1/3, etc. so what are you left with?

    I believe the original post gave you 2 suggestions:
    1) Use partial fractions.
    2) Use properties of telescoping series.

    1) was done for you by the Russian devil, now do part 2)
  9. Dec 29, 2008 #8
    Yes, they will cancel out and the series simply equal 1 therefore the series converge to one. I'll go & try some more examples & hopefully I think I learned the idea of "telescoping series"!

    Thanks very much :)
  10. Dec 30, 2008 #9
    No, look again.

    There are [itex]1 + \frac{1}{n+1}[/itex] left.

    Now you got [tex]\lim_{n \rightarrow \infty}(1 + \frac{1}{n+1})[/tex].

    Because [tex]\lim_{n \rightarrow \infty}(\frac{1}{n+1})=0[/tex]

    you got what? :smile: (rhyme)
    Last edited: Dec 30, 2008
  11. Dec 30, 2008 #10
    Hold on here for a moment!


    In the original post i can see that the series is actually


    and not


    There is a difference, don't u think? There certainly is a difference, unless the OP actually wanted to write the second one.

    (To the OP:which one is it?)
  12. Dec 30, 2008 #11
    He meant . as a a multiplication symbol, what he probably should have used was \cdot
  13. Dec 30, 2008 #12
    Well, yeah, that's what i was thinking too, since it isn't that easy to evaluate the sum of the other seires, unless we sought for some relation to a specific case of a Taylor expansion, or something like that.
  14. Jan 19, 2009 #13
    Hi StupidMath! Yes, like NomoreExams said I meant a multiplication symbol.

    Sorry Dyavol, I forgot.

    [tex]lim_{n\rightarrow \infty} (1 - \frac{1}{n+1}) = 1 - 0 = 1[/tex]

    Hence the series converges and its sum is 1.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Convergence for Infinite Series