Question

Looking at the sum from 1 to INF of 8/(n^2+2n). So far, I've broken it down into the partial fraction so that the sum is for [(4/n)-(4/n+2)]. From there I treated it as two separate sums. I'll attached my nth attempt. :)

Answer 1

I meant to say "I'll *attach*. :P Anyhow, here it is.

Answer 2

Mmmm ok I like your work up to this point,
\[\Large\rm \sum_1^{\infty}\frac{4}{n}-\sum_1^{\infty}\frac{4}{n+2}\]When I first looked at it I got confused for a minute also.
This appears to be a p-series with p=1, yes?
NOT a geometric series.

No power is being applied to our 4's.

Answer 3

Eeeks. This is going to take a while before it starts coming naturally.

Answer 4

Ok, I think I can work with that.

Answer 5

I'm a little rusty on my power series I'm afraid :(
But Wolfram is saying this `converges` by the comparison test.
So maybe you can try working with the harmonic series or something...

Answer 6

Yeah, I can't get the 4/n to converge, so I guess I'd better brush up on harmonic series.

Answer 7

Well we have a `divergent series` minus a `divergent series` which isn't really telling us much....
Maybe we didn't want to do partial fractions.
Maybe there is something we can compare it to from it's initial form.
Hmm

Answer 8

I tried pulling the 8 out in front, but I don't really think that'll buy me anything.

Answer 9

Could I take an integral from 1 to INF ?

Answer 10

We could do this I guess,
By the comparison test,\[\Large\rm 8\sum_{n=1}^{\infty}\frac{1}{n^2+2n}\quad \lt \quad 8\sum_{n=1}^{\infty}\frac{1}{n^2}\]Since we know that the sum on the right converges, we know that the one on the left will.
See how every term will be a little bit smaller (in the left series) because of the +2n in the `denominator` ?

This doesn't tell us `what it converges to`, but it least it's a start :(

Answer 11

Hmm... so Wolfram does show it converging to 6, but they don't show how. :P
I'll have to sit on this for a while. I'm headed to a family gathering tonight. My bro-in-law just finished his math BS, maybe he'll be able to shed some insight, too. I'll check in after. Thanks again for all your help.

Answer 12

math BS? Ooo nice! :)

Answer 13

Math is totally BS sometimes lol

Answer 14

Ha! I think I'm on the right path now. Really, this is just asking us for the area under the curve, right? So if I take the integral from 1 to INF of the function, and the limit as t -> INF of the answer, that should do it.

Answer 15

Hint: telescoping series.

Answer 16

\[\sum_{n=1}^{\infty}\frac{4}{n}-\sum_{n=1}^{\infty}\frac{4}{n+2}=4\left(1+\frac{1}{2}+\frac{1}{3}+\cdots\right)-4\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots\right)\]
Terms will disappear and the sum will "telescope"

Answer 17

Ohhh I totally missed that >.< hah!

Answer 18

D'oh! Wow. It sure is a lot easier when you do it right. :) Thanks everyone.