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Finding the sum of an indefinite series

  1. Mar 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Ʃ(x-->infinity, x>0) 11/(n(n+2))


    2. Relevant equations



    3. The attempt at a solution

    I am not quite sure what to do, i think that i am supposed to put it into partial fractions.


    i changed it to the form 11/(n^2+2n)
    --> 11/((n+1)(n+1)-1))
    --> 11/((n+1)^2-1)

    Now i am kind of stuck, i know the answer is 33/4 but i am not sure how i can get to it.

    Can anyone please help me out?

    Also, can this be considered a geometric series?
     
  2. jcsd
  3. Mar 7, 2013 #2

    Dick

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    No, it's not geometric. You would be right to put it into partial fractions, why don't you do that? Partial fractions means expressing it as A/n+B/(n+2). What are A and B?
     
  4. Mar 7, 2013 #3

    LCKurtz

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    No, it isn't a geometric series. Forget the 11 for a moment. Write$$
    \frac 1 {n(n+2)}= \frac A n + \frac B {n+2}$$Figure out ##A## and ##B##, then write out a bunch of terms of$$
    \sum_{n=1}^\infty \left( \frac A n + \frac B {n+2}\right)$$and you will see what it converges to. Then multiply by your ##11##.
     
  5. Mar 7, 2013 #4
    I am unsure of how to take partial fractions using 11/((n+1)^2-1)

    i know that if you have multiplication at the bottom (ie 2 brackets) i could seperate them and multiply each top term by the denominator of the other fraction, and then equate the 2 top terms to equal 11. I am not sure how i would treat this when i have subtraction in the denominator, should i not have changed it into this form?
     
  6. Mar 7, 2013 #5

    Dick

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    You haven't changed it in any useful way. Just deal with 1/(n(n+2)) and do partial fractions on that. You can save the 11 till later.
     
  7. Mar 7, 2013 #6
    ok, unless i've made calculation errors ive found A to be 1/2 and B to be -1/2. The first 3 terms of the series without the 11 in the numerator i've found are 1/3, 1/8, and 1/15, and 1/24. It seems like they are converging towards 0, but i don't think they ever reach 0, so how do i know what term to stop at?
     
  8. Mar 7, 2013 #7

    Dick

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    Don't combine the parts coming from partial fractions. That gets you right back where you started. (1/2-1/6)+(1/4-1/8)+(1/6-1/10)+(1/8-1/12)+(1/10-1/14)+(1/12-1/16)+... Look for a pattern of cancellations. It's a telescoping series.
     
  9. Mar 7, 2013 #8
    i can see the cancellations now, but isn't this still an infinite series? even if every few terms cancel out how can you tell when you have to keep going until?
     
  10. Mar 7, 2013 #9

    Dick

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    They ALL cancel except for a few at the beginning, the series is infinite. Just find the ones that don't cancel with later terms. Like 1/2 will never cancel. Keep thinking about it.
     
  11. Mar 7, 2013 #10
    ohh i see now.

    Thanks for all the help
     
  12. Mar 8, 2013 #11

    Ray Vickson

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    By definition, an infinite series is the limit of a finite series, so
    [tex] \sum_{n=1}^{\infty} \frac{1}{n(n+2)}=
    \lim_{N \to \infty} \sum_{n=1}^{N} \frac{1}{n(n+2)},[/tex]
    and for a 'telescoping' series the finite sum is easy to get.
     
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