Finding the formula for the partial sum Sn

[chris4642]

Homework Statement

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


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
S_n=? I can't find any ratio or exponential relation between the terms. I need to find a general formula for S_n

 

[Infrared]

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

 

[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?

 

[LCKurtz]

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?

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.

 

[ehild]

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

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

ehild

 

[Curious3141]

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?

Have you learned about partial fractions?

 

[HallsofIvy]

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?