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Series convergence

  1. Jun 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Use partial fractions to show
    [tex] \displaystyle\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)} = \frac{1}{4}[/tex]

    3. The attempt at a solution
    I did the partial fraction decomposition to get: [tex] \displaystyle\sum_{n=1}^\infty \frac{1}{2n} - \frac{1}{n + 1} + \frac{1}{2n + 4} [/tex]

    I'm not sure how to proceed from here, in my textbook the example shows how terms in the partials sums overlap and cancel out if you start looking at the terms in the partial fraction decomposition, however I can't see that happening with this particular series.

    Any suggestions would be appreciated. Thanks in advance.
     
  2. jcsd
  3. Jun 10, 2009 #2

    rock.freak667

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    Write out a few term, n=1,2,3,4,5....N-3,N-2,N-1,N

    then check as N->inf.
     
  4. Jun 10, 2009 #3
    Hi,

    I tried doing that originally, but I don't see any patterns that I can exploit when I write the terms of the partial decomposition for n = 1, 2, 3,..., N-2, N-1, N as you stated.

    Do you have any further suggestions?
     
  5. Jun 10, 2009 #4

    Dick

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    Write the terms for successive values of n each on a row 1/(2n+4) -1/(n+1) 1/(2n)
    1/6 -1/2 1/2
    1/8 -1/3 1/4
    1/10 -1/4 1/6
    1/12 -1/5 1/8
    1/14 -1/6 1/10

    Oblique hint: diagonal.
     
  6. Jun 10, 2009 #5
    Interesting, so everything seems to be canceling out if you keep going long enough diagonally, except the 1/4 in the second row. However is there any way to formalize this? Or is simply noticing this pattern enough for a formal proof?
     
  7. Jun 10, 2009 #6

    Dick

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    Depends on how formal you want to be. Writing a table like that and scratching out the cancellations is good for me. If you want to do it formally write out the first term from the n case, the second term from the n+1 case and the third term from the n+2 case and show they cancel algebraically. What is important is to realize just because there isn't an obvious cancellation doesn't mean there isn't one.
     
  8. Jun 10, 2009 #7
    Thank you for the help!
     
  9. Jun 10, 2009 #8

    txy

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    why not take out 1/2 from the partial fraction decomposition so that it's more obvious?

    [tex]\sum^{+\infty}_{n=1}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})[/tex]
    [tex]= \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2})[/tex]
    [tex]= \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}-\frac{1}{n+1})[/tex]
    [tex]= \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n}-\frac{1}{n+1}) + \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n+2}-\frac{1}{n+1})
    [/tex]
     
  10. Jun 10, 2009 #9

    Dick

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    That's a nice way to handle it. There's more than one way to skin a cat.
     
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