Question

Series : converges or not : ((cos n)^2)/(n^2 + 1) ? and how you do it ?

Answer 1

Use the squeeze theorem. Since:\[0\le \left(\cos n \right)^2\le 1\]Dividing everything by n^2+1 yields:\[0\le \frac{(\cos n )^2}{n^2+1}\le \frac{1}{n^2+1}\]What is the limit of the rightmost term as n goes to infinity? What is the limit of the leftmost term? What can you conclude about the middle term?

Answer 2

but the left one is not -1 / (n^2 +1) ?

Answer 3

Generally that is the way to go, but the square around the cos n term makes it positive again.

Answer 4

ooo alright thx alot !

Answer 5

the limit is 0 ?

Answer 6

Yes thats correct :)

Answer 7

but converges or not because general term say if = 0, we cant conclude anything

Answer 8

Ah, I didnt realize it was a question about a series, and not a sequence, my apologies.

Answer 9

So you are correct, that test doesn't conclude anything about the series. So we should use a direct comparison.

Answer 10

Note that:\[\frac{(\cos n )^2}{n^2+1}\le \frac{1}{n^2+1}\le \frac{1}{n^2}\]

Answer 11

P series ? \[n > 1\] so converges ?

Answer 12

Thats correct, so since \[\sum_{n=1}^{\infty}\frac{1}{n^2}\]converges, and:\[\frac{(\cos n)^2}{n^2+1}\le \frac{1}{n^2}\], what can we conclude about:\[\sum_{n=1}^{\infty}\frac{(\cos n)^2}{n^2+1}\]?

Answer 13

it converges !