# Sum of Infinite series, does it converge?

1. Jan 16, 2007

### rcmango

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

Find the sum of the infinite series, if the series converges.

infinity
E
n = 1

(2) / (n^2 + 2n)

2. Relevant equations

..

3. The attempt at a solution

I believe this problem doesn't look very hard, I think all i really need to do is divide the denominator and numerator by the largest n in the denominator.

so it would look like this --> (2/n^2) / (1 + 2/n)

which would become 0/1

so converges to 0?

but what is finding the sum of the infinite series?

2. Jan 16, 2007

### StatusX

You've shown that the terms go to zero as n->infinity. If this isn't true, the series couldn't converge, but by itself it doesn't guarantee the series converges. For example, even though $1/n \rightarrow 0$:

$$\sum_{n=1}^\infty \frac{1}{n} = \infty$$

Are you familiar with the series:

$$\sum_{n=1}^\infty \frac{1}{n^2}$$

? You can use this to show that your series converges. To find the sum, try to make it into a telescoping series.

3. Jan 16, 2007

### rcmango

i'm not too familiar with all the different series quite yet.

i'm confused when they say 'find the sum'

Am i trying to prove that this problem is similiar to a series pattern?
if so, we're on track to use the telescoping series to do this.

thankyou.

4. Jan 16, 2007

### HallsofIvy

Staff Emeritus
Interesting that you should mention "telescoping series"! Use "partial fractions" to write $\frac{2}{n^2+ 2n}$ as difference of two fractions and see what you get.

5. Jan 16, 2007

### rcmango

ya someone up above recommended the telescoping series..

for partial fractions I tried using a formula, but i don't think i reached the correct partial fractions.

first i factored this fraction: 2/(n)(n+2)

then used substitution to reach factorials, but i don't think i reached the correct answer --> 2/n(n+2) = 1/n + 1/(n+2)

are we making progress?
thanks so far.

6. Jan 17, 2007

### dextercioby

It's with a minus

$$\frac{2}{n(n+2)}=\frac{1}{n}-\frac{1}{n+2}$$

Daniel.