Question

Determine if the sequence (cosn/ n) converges or diverges. tell if it is absolutely or conditionally convergent

Answer 1

http://www.wolframalpha.com/input/?i=summation+from+0+to+infinity+cos+x%2Fx

does not converge

Answer 2

i forgot to write this down sorry but n=1 not 0

Answer 3

\[ \cos x = 1 - x^2/2! + x^4/4! - x^6/6!+ .. = \sum_{i}^{n} (-1)^i x^{2i}/(2i)!\]
\[ \large \sum_{n=0}^{\infty} \frac{ \sum_{i}^{\infty } (-1)^i n^{2i}/(2i)!}{n}\]1

Answer 4

but we do have fixed value for n=1, isn't our series going to be like this??

don't worry, i don't know how to solve it. LOL

Answer 5

would squeeze theorem work
−1≤cosn≤1 and since −1/n≤cosn/n≤1/n
finding the limit of each of them and if they all are 0 then it converges right

Answer 6

1/n is divergent itself ... and cosn/n <= 1/n -> gives no info on it.

Answer 7

o ... i was putting zero at beginning

Answer 8

it converges.
http://www.wolframalpha.com/input/?i=summation+from+1+to+infinity+cos+x%2F+x

Answer 9

so would you say it is conditionally or absolutely convergent?

Answer 10

it says it converges conditionally ... since it did not put absolute value
also converges absolutely
http://www.wolframalpha.com/input/?i=summation+from+1+to+infinity+abs%28+cos+x%2F+x+%29

Answer 11

Couldn't you do a limit comparison with 1/n? If the limit exists, both diverge or both converge. It simplifies to the limit of |cos(n)|, but I don't think the limit exists, right? Another way was to try to compare, maybe.

Answer 12

by comparison test since cosn/n <= 1/n is convergent
and 1/n is divergent by harmonic series
so the series is conditionally converget
is that right?

Answer 13

time for dinner .. BRB
@eliassaab we require your help

Answer 14

No. If {bn} is divergent, then {an} has to be greater to you come to a conclusion. {an} is smaller, no conclusion can be reached.