Finding the formula for the partial sum Sn

1. Jul 27, 2013

chris4642

1. The problem statement, all variables and given/known data

Consider the series Ʃ 1/[k(k+2)]; n=1 to infinity
Find the formula for the partial sum Sn

2. The attempt at a solution
I have calculated the first 5 terms of the sequence as follows, but I can't see any pattern. Am I doing this right?
S1=1/3
S2=1/3+1/8=11/24
S3=1/3+1/8+1/15=21/40
S4=1/3+1/8+1/15+1/24=17/30
S5=1/3+1/8+1/15+1/24+1/35=25/42
Sn=??? I can't find any ratio or exponential relation between the terms. I need to find a general formula for Sn
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 27, 2013

Infrared

Hint: $\frac{1}{k(k+2)}=\frac{1}{k}-\frac{1}{k+2}$. Your series is telescoping.

3. Jul 27, 2013

chris4642

Telescoping series

Ok I have read a bit on telescoping series. One thing I do not understand is this:

Why does it go from 1/[k(k+2)] to 1/k - 1/k+2, why are they subtracted instead of added?

4. Jul 27, 2013

LCKurtz

What happens if you simplify $\frac 1 k - \frac 1 {k+2}$? Does it equal $\frac 1 {k(k+2)}$ or not? That will answer your question.

5. Jul 28, 2013

ehild

It is $\frac{1}{k(k+2)}=0.5\left(\frac{1}{k}-\frac{1}{k+2}\right)$

ehild

6. Jul 28, 2013

Curious3141

Have you learnt about partial fractions?

7. Jul 28, 2013

HallsofIvy

Staff Emeritus
More to the point, don't you know how to add and subtract fractions?
To add or subtract 1/k and 1/(k+2) (the parentheses in that second fraction are important!) you need to get the same denominator which is, of course, k(k+2).

1/k+ 1/(k+2)= (k+2)/k(k+2)+ k/k(k+2)= (2k+2)/k(k+2)

1/k- 1/(k+2)= (k+2)/k(k+2)- k/k(k+2)= 2/k(k+2)

Which is the one you want?