# Homework Help: Use partial fractions to find the sum of the series

1. Aug 12, 2012

### JFUNR

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

2. Relevant equations
Partial Fraction breakdown: $\Sigma$ $\frac{5}{2n}$+$\frac{5}{2(n+2)}$+$\frac{5}{(n+1)}$

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. Aug 12, 2012

### Ray Vickson

Your partial fraction expansion is incorrect.

RGV

3. Aug 12, 2012

### JFUNR

ah that was a typo, the partial fraction expansion I have in my attempts is...$\frac{5}{2n}$+$\frac{5}{2(n+2)}$-$\frac{5}{n+1}$

I still have the same issue..

4. Aug 12, 2012

### Zondrina

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
5. Aug 12, 2012

### gabbagabbahey

Hint 1: $$\sum_{n=1}^{\infty} f(n) = \lim_{N \to \infty} \sum_{n=1}^{N} f(n)$$

Hint 2: $$\sum_{n=1}^{N} \frac{1}{n+2} = \left( \sum_{n=1}^{N+2} \frac{1}{n} \right) - \frac{1}{1} - \frac{1}{2}$$

6. Aug 12, 2012

### qbert

reindex second and third sums.
and notice you get a whole lot of cancelation.

7. Aug 13, 2012

### JFUNR

How do you mean?

8. Aug 13, 2012

### Ray Vickson

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

RGV

9. Aug 13, 2012

### JFUNR

by all means...enlighten me..where's the pattern........

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

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
10. Aug 13, 2012

### Ray Vickson

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

RGV

11. Aug 14, 2012

### gabbagabbahey

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

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.