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Sum of infinite series

  1. Feb 27, 2005 #1
    hey, i was just wondering if anyone could help find the sum of the infinite series defined by 1/[n(n+1)(n+2)]. I can split it into partial fractions but not sure from there. Thanks
     
  2. jcsd
  3. Feb 27, 2005 #2
    its a/(1-r)...
    where |r| < 1

    where a is the initial values (1/6), and r is the common ratio...
    it to late for my brain to think right now so I can tthink of what the ratio is going to be.. wil try tomorrow if nobody else has helped you.. going tp sleep now..gonight!
     
  4. Feb 27, 2005 #3

    dextercioby

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    I'm not sure the splitting can do you any good,because each of the series you'd be geting would be infinite.If you don't know the answer,well,my Maple said it's 1/4...i don't know how it did it...

    I'm gonna check it in G & R,too...

    Daniel.
     
  5. Feb 27, 2005 #4
    i thought you could do it by splitting it into partial fractions, then finding an expression for the nth partial sum and let n tend to infinity
     
  6. Feb 27, 2005 #5

    dextercioby

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    According to Maple,the partial sum is:
    [tex] \sum_{k=1}^{n}\frac{1}{k(k+1)(k+2)} =-\frac{1}{2(n+2)(n+1)}+\frac{1}{4} [/tex]

    If you can find a way to prove what i've just written,then that's it...

    Daniel.
     
  7. Feb 27, 2005 #6

    Hurkyl

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    I think partial fractions is promising too -- with care, you can probably arrange the terms into one or more telescoping series. What did you get when you applied partial fractions?
     
  8. Feb 27, 2005 #7

    dextercioby

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    Here's the proof for the partial sum:

    [tex] S=:\sum_{k=1}^{n}\frac{1}{k(k+1)(k+2)}=\frac{1}{4}-\frac{1}{2(n+1)(n+2)} [/tex](1)
    -----------------------------||------------------------

    Use partial fractions to rewrite the initial partial sum as:

    [tex] S=\sum_{k=1}^{n}\frac{1}{k(k+1)(k+2)}=\sum_{k=1}^{n}\frac{1}{2k}-\sum_{k=1}^{n}\frac{1}{k+1}+\sum_{k=1}^{n}\frac{1}{2(k+2)} [/tex] (2)

    According to Maple each of the three sums is:

    [tex] \sum_{k=1}^{n}\frac{1}{2k}=\frac{1}{2}\psi(n+1)+\frac{1}{2}\gamma [/tex] (3)

    [tex] \sum_{k=1}^{n} \frac{-1}{k+1}=-\psi(n+2)+1-\gamma [/tex] (4)

    [tex] \sum_{k=2}^{n} \frac{1}{2(k+2)}=\frac{1}{2}\psi(n+3)-\frac{3}{4}+\frac{1}{2}\gamma [/tex](5)

    Add (3)->(5) and equate with (2):

    [tex] S=\frac {1}{4}+\frac{1}{2}\psi(n+1)-\psi(n+2)+\frac{1}{2}\psi(n+3) [/tex](6)

    Now,use the property of "psi":

    [tex] \psi(z+1)=\psi(z)+\frac{1}{z} [/tex] (7) for [itex]z\neq 0 [/itex]

    to write:

    [tex] \psi(n+2)=\psi(n+1)+\frac{1}{n+1} [/tex] (8)

    [tex] \psi(n+3)=\psi(n+2)+\frac{1}{n+2}=\psi(n+1)+\frac{1}{n+1}+\frac{1}{n+2} [/tex] (9)

    to rewrite the "3-psi" term from (6) simply as:

    [tex] -\frac{1}{2(n+1)(n+2)} [/tex] (10)

    Then add (10) to 1/4 from (6) to get the equality you were supposed to prove... :wink:

    In the initial equality set [itex] n\rightarrow +\infty [/itex] to get the answer [itex] \frac{1}{4} [/itex]

    Daniel.
     
  9. Feb 27, 2005 #8

    Hurkyl

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    What do you think of this decomposition?

    [tex]
    \frac{1}{k (k+1) (k+2)} = \frac{1}{2}
    \left(
    \frac{1}{k} - \frac{1}{k+1} - \frac{1}{k+1} + \frac{1}{k+2}
    \right)
    [/tex]
     
  10. Feb 27, 2005 #9

    dextercioby

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    That one gives the answer,too...1/4...

    Daniel.
     
  11. Feb 28, 2005 #10

    i got this far, then not sure how to cancel terms to find the sum?
     
  12. Feb 28, 2005 #11

    dextercioby

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    Well,it's not difficult.You may write it as
    [tex] \frac{1}{2}[(\frac{1}{k}-\frac{1}{k+1})-(\frac{1}{k+1}-\frac{1}{k+2})] [/tex]

    and then give values to "k" to see the pattern the terms take...

    Daniel.
     
  13. Feb 28, 2005 #12
    yeah got it now, thanks for all the help!
     
  14. Feb 28, 2005 #13

    saltydog

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    Thanks, didn't have a clue. Turns out the first one is the following:

    [tex]\sum_{k=1}^{n}\frac{1}{2k}=\frac{1}{2}\{\frac{\int_0^\infty e^{-t}t^{n}\ln(t)dt}{\int_0^\infty e^{-t}t^{n}dt}+\int_0^\infty \frac{1}{t}(\frac{1}{1+t}-e^{-t})dt\} [/tex]

    I had no idea! Does Adamg know this? Don't mind me, I mean I had problems with Leibnitz's rule the other day. I tell you what though, if I was the teacher assigning this problem, we'd be looking at these integrals. Now how in the world does the sum turn out to be this messy expression?
     
  15. Mar 1, 2005 #14

    dextercioby

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    It comes from the logarithmic derivative of the Gamma Euler function,which is "psi"...

    Daniel.
     
  16. Nov 2, 2009 #15
    Is an easy sum:

    [tex]\sum\frac{1}{n(n+1)(n+2)}[/tex]=[tex]\sum\frac{1}{2n}[/tex]+[tex]\frac{1}{2(n+2)}[/tex]-[tex]\frac{1}{n+1}[/tex]= ([tex]\frac{1}{2}[/tex]+[tex]\frac{1}{4}[/tex]+[tex]\frac{1}{6}[/tex]+[tex]\frac{1}{8}[/tex]+[tex]\frac{1}{10}[/tex]...+[tex]\frac{1}{6}[/tex]+[tex]\frac{1}{8}[/tex]+[tex]\frac{1}{10}[/tex]+...+(-[tex]\frac{1}{2}[/tex]-[tex]\frac{1}{3}[/tex]-[tex]\frac{1}{4}[/tex]-[tex]\frac{1}{5}[/tex]-[tex]\frac{1}{6}[/tex]-...) = [tex]\frac{1}{2}[/tex]+[tex]\frac{1}{4}[/tex]+[tex]\frac{1}{3}[/tex]+[tex]\frac{1}{4}[/tex]+[tex]\frac{1}{5}[/tex]+.... +(-[tex]\frac{1}{2}[/tex]-[tex]\frac{1}{3}[/tex]-[tex]\frac{1}{4}[/tex]-[tex]\frac{1}{5}[/tex]-[tex]\frac{1}{6}[/tex]-...) = [tex]\frac{1}{4}[/tex].
     
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