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Use partial fractions to find the sum of the series

  1. Aug 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Use partial fractions to find the sum of the series: [itex]\Sigma[/itex]n=1 to infinity [itex]\frac{5}{n(n+1)(n+2}[/itex]


    2. Relevant equations
    Partial Fraction breakdown: [itex]\Sigma[/itex] [itex]\frac{5}{2n}[/itex]+[itex]\frac{5}{2(n+2)}[/itex]+[itex]\frac{5}{(n+1)}[/itex]


    3. The attempt at a solution
    When I tried to cancel terms out, it is erratic and no pattern seems to emerge. It is not geometric so using a/(1-r) isn't possible. When I separated the terms, each diverges based on the nth term test. According to the all-knowing wolfram-alpha, the series converges to 5/4...I feel as if I am missing something big in my attempts at solving this one? any help?
     
  2. jcsd
  3. Aug 12, 2012 #2

    Ray Vickson

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    Your partial fraction expansion is incorrect.

    RGV
     
  4. Aug 12, 2012 #3
    ah that was a typo, the partial fraction expansion I have in my attempts is...[itex]\frac{5}{2n}[/itex]+[itex]\frac{5}{2(n+2)}[/itex]-[itex]\frac{5}{n+1}[/itex]

    I still have the same issue..
     
  5. Aug 12, 2012 #4

    Zondrina

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    Edit : Whoops you had the right expansion I didn't see your post there.

    There will be a pattern you will notice, but a more important question to ask yourself is : Suppose your sum is going from 1 to infinity. Lets say you're summing some A+B+C which are all composed of a variable n ( the variable you are summing of course ). If one of A, B, or C diverges, does the whole series diverge?
     
    Last edited: Aug 12, 2012
  6. Aug 12, 2012 #5

    gabbagabbahey

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    Hint 1: [tex] \sum_{n=1}^{\infty} f(n) = \lim_{N \to \infty} \sum_{n=1}^{N} f(n)[/tex]

    Hint 2: [tex]\sum_{n=1}^{N} \frac{1}{n+2} = \left( \sum_{n=1}^{N+2} \frac{1}{n} \right) - \frac{1}{1} - \frac{1}{2}[/tex]
     
  7. Aug 12, 2012 #6
    reindex second and third sums.
    and notice you get a whole lot of cancelation.
     
  8. Aug 13, 2012 #7
    How do you mean?
     
  9. Aug 13, 2012 #8

    Ray Vickson

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    He means that you should write it out in detail and see what happens.

    RGV
     
  10. Aug 13, 2012 #9
    by all means...enlighten me..where's the pattern........

    ([itex]\frac{5}{2}[/itex]+[itex]\frac{5}{6}[/itex]-[itex]\frac{5}{2}[/itex])+([itex]\frac{5}{4}[/itex]+[itex]\frac{5}{8}[/itex]-[itex]\frac{5}{3}[/itex])+([itex]\frac{5}{6}[/itex]+[itex]\frac{5}{10}[/itex]-[itex]\frac{5}{4}[/itex])+([itex]\frac{5}{8}[/itex]+[itex]\frac{5}{12}[/itex]-[itex]\frac{5}{5}[/itex])+([itex]\frac{5}{10}[/itex]+[itex]\frac{5}{14}[/itex]-[itex]\frac{5}{6}[/itex])+([itex]\frac{5}{12}[/itex]+[itex]\frac{5}{16}[/itex]-[itex]\frac{5}{7}[/itex])+([itex]\frac{5}{14}[/itex]+[itex]\frac{5}{18}[/itex]-[itex]\frac{5}{8}[/itex])+...+

    while some terms do cancel, I do not see a patter out of the first 7 terms...but like I said...enlighten me
     
    Last edited: Aug 13, 2012
  11. Aug 13, 2012 #10

    Ray Vickson

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    Do you really need someone to tell you that
    [tex] \sum_{k=1}^N \frac{1}{2k} = \frac{1}{2} \sum_{k=1}^N \frac{1}{k}\;?[/tex]

    RGV
     
  12. Aug 14, 2012 #11

    gabbagabbahey

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    The pattern may indeed be a little difficult to see like this, but notice, for example, that [itex]\frac{5}{6}+\frac{5}{6}-\frac{5}{3}=0[/itex] and [itex]\frac{5}{8}+\frac{5}{8}-\frac{5}{4}=0[/itex].

    To formally reorganize the sum so that the cancellations become obvious, first look at the finite sum and apply the second hint I gave you to both the 5/(2(n+2)) and -5(n+1) terms. After reorganizing the finite sum to get the desired cancellations, take the limit as N->infinity.
     
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