What is the sum of this series?

[utkarshakash]

Homework Statement

If f(r) = r^3-5r^2+6r then \sum_{r=4}^{\infty } \dfrac{1}{f(r)} is

The Attempt at a Solution

I could decompose the above summation into something like this

\dfrac{1}{3} \left( \sum \dfrac{1}{(r-2)(r-3)} - \sum \dfrac{1}{r(r-2)} \right)

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.

:(

 

[epenguin]

Homework Statement

If f(r) = r^3-5r^2+6r then \sum_{r=4}^{\infty } \dfrac{1}{f(r)} is

The Attempt at a Solution

I could decompose the above summation into something like this

\dfrac{1}{3} \left( \sum \dfrac{1}{(r-2)(r-3)} - \sum \dfrac{1}{r(r-2)} \right)

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.

:(

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?

 

[LCKurtz]

Homework Statement

If f(r) = r^3-5r^2+6r then \sum_{r=4}^{\infty } \dfrac{1}{f(r)} is

The Attempt at a Solution

I could decompose the above summation into something like this

\dfrac{1}{3} \left( \sum \dfrac{1}{(r-2)(r-3)} - \sum \dfrac{1}{r(r-2)} \right)

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.

:(

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.

 

[HallsofIvy]

"Partial fraction" tells you that you can write
\frac{1}{r^3- 5r^2+ 6r}= \frac{A}{r}+ \frac{B}{r- 2}+ \frac{C}{r- 3}
for constants, A, B, and C. And that gives a "telescoping series".

 

[Saitama]

"Partial fraction" tells you that you can write
\frac{1}{r^3- 5r^2+ 6r}= \frac{A}{r}+ \frac{B}{r- 2}+ \frac{C}{r- 3}
for constants, A, B, and C. And that gives a "telescoping series".

But that isn't a good idea in my opinion, it would be best to follow LCKurtz advice. :)