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Series (help!)

  1. Apr 7, 2004 #1
    Could anybody please tell me how to do this?

    Σn/[(n+1)(n+2)(n+3)]

    Thanks!
     
  2. jcsd
  3. Apr 7, 2004 #2

    Zurtex

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    You tried splitting up into partial fractions and using the method where you have one function minus another?
     
  4. Apr 7, 2004 #3
    I've already splitted it. It's interesting that each fraction alone diverges but the whole sum converges. But now I don't know what to do. What is this method you told me about?

    Thanks
     
  5. Apr 7, 2004 #4
    as zurtx said you can split it like this : Σ(1/[n+3])*(1/[n+1]-1/[n+2])=1/[n+3]*Σ1/[n+1]-1/[n+2]).
     
  6. Apr 7, 2004 #5
    Still can't solve it. My computer says it certainly converges to 1/4. How can I prove that?
     
  7. Apr 7, 2004 #6
    did you learned the term when a series is telescoped (or something of that that sort i have it on a book in hebrew in level of calclus 2)?
     
  8. Apr 7, 2004 #7

    matt grime

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    LQG, you can't take that term outside the summation as the n is part of the summation index.
     
  9. Apr 7, 2004 #8
    it cannot be representing a coefficient in the form of Σd*(a1-an)
    where d=1/(n+3) is the coefficient?
     
  10. Apr 7, 2004 #9

    matt grime

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    Apologies for getting LGQ and <acronym title='Loop Quantum Gravity' style='cursor:help;'>LGQ</acronym> mixed up (if it's any consolation I'm reading a paper on TQFT and such as I write (well, not exactly "as I write" but you get the idea))

    You're taking the sum indexed by n when you say 1/[n+3]*Σ1/[n+1]-1/[n+2] ie n runs from something to something. You can't take the n+3 outside like that.

    It's just like taking the x outside an integral wrt x (of course it's possible that it is the interior part of the summation that is wrong)
     
    Last edited: Apr 7, 2004
  11. Apr 7, 2004 #10
    matt grime is right. Yet that doesn't solve the problem. Thanks anyway for your help
     
  12. Apr 7, 2004 #11

    matt grime

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    but you can do it by separating into partial fractions, it's quite easy isn't it?
     
  13. Apr 7, 2004 #12

    Zurtex

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    Where A, B, C are arbitrary constants let:

    [tex]\frac{n}{(n+1)(n+2)(n+3)} = \frac{A}{n+1} + \frac{B}{n+2} + \frac{C}{n+3}[/tex]

    Work out A, B and C. Now I could be wrong but I got A and C as negative numbers. So you should be able to write out the functions first few values n= 1, 2, 3, 4 and spot a pattern that if you add them together terms will start to cancel and you can extend that to n = r-3, r-2, r-1, r and get a general equation for the sum of f(n) from n = 1 to r.
     
  14. Apr 8, 2004 #13
    I just worked this out and the answer is in fact 1/4. Zurtex has the right approach. Find A, B and C and then write out the first several terms for each partial fraction. You should see a bunch of them cancel and you'll be left with a couple terms per fraction. From there, you can see the limit.
     
  15. Apr 9, 2004 #14
    Thanks everybody! You're all great.
     
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