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Find sum of this series

  1. Mar 24, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    If [itex]f(r) = r^3-5r^2+6r [/itex] then [itex]\sum_{r=4}^{\infty } \dfrac{1}{f(r)} [/itex] is


    3. The attempt at a solution

    I could decompose the above summation into something like this

    [itex]\dfrac{1}{3} \left( \sum \dfrac{1}{(r-2)(r-3)} - \sum \dfrac{1}{r(r-2)} \right)[/itex]

    But from here I'm not sure how to take it ahead. I tried putting some values of r to check if they cancel out but to my dismay, they do not.

    :(
     
  2. jcsd
  3. Mar 24, 2014 #2

    epenguin

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    You have noticed the quadratic factorises easily. And it is in a denominator. You have never had to deal with, had any other exercises with a factorised polynomial in a denominator?
     
  4. Mar 24, 2014 #3

    LCKurtz

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    Have you studied telescoping series? Use partial fractions to write$$
    \frac1 {(r-2)(r-3)}= \frac {-1}{r-2}+ \frac 1 {r-3}$$Write out several terms starting with ##r=4## and you will see why it is called telescoping. (Don't simplify by adding the fractions, just leave the fractions as they are and look for cancellations). Similarly with the second summation.
     
    Last edited: Mar 24, 2014
  5. Mar 25, 2014 #4

    HallsofIvy

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    "Partial fraction" tells you that you can write
    [tex]\frac{1}{r^3- 5r^2+ 6r}= \frac{A}{r}+ \frac{B}{r- 2}+ \frac{C}{r- 3}[/tex]
    for constants, A, B, and C. And that gives a "telescoping series".
     
  6. Mar 25, 2014 #5
    But that isn't a good idea in my opinion, it would be best to follow LCKurtz advice. :)
     
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