Question

Determine whether the following series converges or diverges.

Answer 1

\[\sum_{n=1}^{\infty}\frac{ \sin(n)+\cos(n) }{ n^5}\]

Answer 2

I know I have to use the comparison test but I am not sure what comparison to use.

Answer 3

can you split it into 2 parts?

Answer 4

since sin and cos flucuate between -1 and 1; the larger n gets the closer this gets to zero, try to compare it with 1/n^5 maybe?

Answer 5

Erm... Not sure if that's valid...

Answer 6

not sure if 1/n^5 is valid?

Answer 7

Well yeah I was trying but but I don't think that's a valid comparison...

Answer 8

Yeah.

Answer 9

that converges ... use, use comparison test.

Answer 10

sin(n)+cos(n)

cos(n) - sin(n) = 0; when n=45

so the largest sin+cos can be is: sqrt(2)

the smallest it can be is 1, or -1 respectively

right?

Answer 11

Yeah I know.

Answer 12

yeah compare it with 1/n^5 ... you know 1/n^5 < 1/n^2 for all n>1 ... the value of of 1/n^2 is well known to converge to pi^2/6, besides you can use integral test.

Answer 13

Lol I know. I am wondering WHY I can use that comparison.

Answer 14

Because it does not feel valid.

Answer 15

why not valid ... this is perfectly valid.

Answer 16

But I can only use the comparison test for sufficiently large values of n for which the term does not change significantly.

Answer 17

Yeah, I know I am being stubborn at this point.

Answer 18

can you give me an example by what you mean?

Answer 19

sin(n) - cos(n) <= 2 use this and p series

Answer 20

Like... for 1/(2n^2+3n+5) I can use the comparison test because for large values of n the 2n^2 dominates so I can treat it as 1/2n^2 .

Answer 21

@zzr0ck3r : We have not learnt power series yet.

Answer 22

I can't do the same thing for sin(n)+cos(n).

Answer 23

use geometric for k != 1 then check 1 by inspection?

Answer 24

1/k^5 is geometric for k> 1

Answer 25

this is even more obvious ... n^5 > n^2 ... so 1/n^5 < 1/n^2, P series argument is better.

Answer 26

I think choosing geometric series also works .. as zzr0cker mentioned.

Answer 27

yeah but... I can't just say for large values of n I can say (sin(n)+cos(n))/n^5 is the same as 1/n^5

Answer 28

do you have the comparison test ?

Answer 29

Geometric series is a good argument too...