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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

    rock.freak667

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    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
    [tex]\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}[/tex]

    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!

    :bugeye:

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

    [tex]\sum_{n=1}^{\infty}\frac{1}{n.(n+1)}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...[/tex]

    and not

    [tex]\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+...[/tex]

    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.
     
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