Question

determine if this sequence converges absolutely, conditionally or diverge?

cos(n)/n^2

Answer 1

@agent0smith do u know how to do this dear?

Answer 2

Erm, well you probably need to prove it with a test (limit test?), but,

as n approaches infinity, cosn will remain between -1 and 1... so you have a small number, divided by n^2. What's this limit equal to?\[\Large \lim_{n \rightarrow \infty}\frac{ 1 }{ n^2 }\]

Answer 3

0....sorry walking dead is on! lol

Answer 4

So it appears to converge absolutely, since the terms approach zero, and the denominator is n^2 (not just n, since 1/n is NOT convergent, 1/n^2 is). I forget what that's called... comparison test i think.

We can basically use the comparison test (and limit test i guess, i really don't know how much proof we need), compare cosn/n^2 to 1/n^2

Answer 5

http://www.wolframalpha.com/input/?i=sum+of+cosn%2Fn%5E2

it has a value; it's convergent. Also http://www.wolframalpha.com/input/?i=sum+of+cosn%2Fn%5E2+convergent%3F

Answer 6

see thats what i thought bc the denominator is n^2 but i wasnt completely sure! thank you!!!

Answer 7

See, you're smart enough to have the right idea! good job :)