Question

I know this series is convergent. How can I compute its sum?

Answer 1

can i compare it to 1/n?

Answer 2

You can't. You can compare it to 1/n^2 though.

Answer 3

the greater the n the less the term. You see that the denominator is at first 3, then 8, then 15...

Answer 4

ohh so then its convergent

Answer 5

since p>1

Answer 6

when series converges that means that the terms are increasing, when series diverges that means the terms decrease

Answer 7

@SolomonZelman that doesn't work. \[\sum_{n=1}^\infty \frac{1}{n}\] is a divergent series.

Convergent means that when you approach infinity you get a finite number. Divergence means when you approach infinity you get an infinite number. Completely different than what you describe.

Answer 8

if i plug in inf i get 0

Answer 9

i thought if p>1 in 1/n^p then the series is conv.

Answer 10

Yes, which is why I brought it up. In that case p=1 and it's divergent. You can compare your sequence to 1/n^2 because if you distribute it out it becomes: \[\sum \frac{1}{n^2+2n}\]

the n^2 term is going to control how it converges/diverges because as you get larger it becomes more important, similar to how L'Hopital's rule works.

So you compare it to 1/n^2 because of this, and you're right, because 2>1 then it converges by the p-series test like you said.

Answer 11

how do i compute its sum

Answer 12

Well it might be a little difficult, I'm not sure.

Answer 13

"how do i compute its sum"

\[\Large \frac{1}{n(n+2)}=\frac{1}{2n}-\frac{1}{2(n+2)}\]

Answer 14

what do i do next?

Answer 15

this is a telescoping sum. look at the first few terms. Then figure out what happens when \(n\to\infty\)

Answer 16

Gah of course, I should have thought of that, good call.

Answer 17

so isn't it just going to be 1/2 + 1/4 - \[- 1/(2(n+2))\]

Answer 18

wait that doesn't work!

Answer 19

Have you guys tried this method?
It's called partial fraction decomposition
\[\frac{ 7 }{ n(n+2) }=\frac{ 7 }{ 2 }\times (\frac{ 1 }{ n }-\frac{ 1 }{ n+2 })\]

Answer 20

So the series become \[\frac{ 7 }{ 2 }\times(\frac{ 1 }{ 1 }-\frac{ 1 }{ 3 }+\frac{ 1 }{ 3 }+\frac{ 1 }{ 5 }-...)=\frac{ 7 }{ 2 }\times \frac{ 1 }{ 1 }=\frac{ 7 }{ 2 }\]

Answer 21

Do you understand this, @nimamiz

Answer 22

ohhh! make sense