Sum of Infinite series, does it converge?

[rcmango]

Homework Statement

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

infinity
E
n = 1

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

Homework Equations

..

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?
help here please. Thankyou.

 

[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.

 

[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 similar to a series pattern?
if so, we're on track to use the telescoping series to do this.

thankyou.

 

[HallsofIvy]

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.

 

[rcmango]

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.

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.

 

[dextercioby]

It's with a minus

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

Daniel.