Question

Hello! How do I figure out if sigma n=1 to infinity cos(n*pi)/n converges or diverges?

Answer 1

maybe convert it to its taylor representation?

Answer 2

\[cos(u)=1+\sum_{k=1}^{\infty}\frac {(-1)^{k}}{(2k)!}u^{2k}\]
\[cos(\pi~n)=1+\sum_{1}^{\infty}\frac {(-1)^{n}}{(2k)!}(\pi)^{2k}~(n)^{2k}\]
\[\frac{cos(\pi~n)}{n}=1+\sum_{1}^{\infty}\frac {(-1)^{n}}{(2k)!}(\pi)^{2k}~(n)^{2k-1}\]
or maybe a ratio test?

Answer 3

my first ideas bad .... need more MtDew this early

Answer 4

hmm our teacher never taught us the Taylor representation, I remember her telling us she wasn't going to.. I'll try the ratio test..

Answer 5

yeah, ratio will help to define the interval of convergence ... at least thats what by addled brain is telling me

Answer 6

sigma n=1 to infinity cos(n*pi)/n :

My first impulse would be to do a comparison test.
If you think about it, you'll likely conclude, as I have, that \[\frac{ \cos n \pi }{ n }\le \frac{ 1 }{ n }\]

Answer 7

Note that the series 1/n is the "harmonic series" and is divergent.
Unfortunately, that doesn't help us much in determining whether or not the given series is convergent or divergent.

Answer 8

alternating series might be useful since we are dealing with an alternating trig

Answer 9

@amistre64 : an alternating series would include a factor such as (-1)^n, which cos (n*pi) does not have.

Aha! inspiration! Would one of you guys please write out the first few terms of the sequence cos n*Pi? let n=0, 1, 2, 3, ... What could you conclude from that?

Answer 10

cos(pi) = -1
cos(2pi)/2 = 1/2
cos(3pi)/3 = -1/3
cos(4pi)/4 = 1/4

Answer 11

so it looks to be an alternating hyperG :)

Answer 12

cos(pi*n) is equal to (-1)^n

Answer 13

That's a significant discovery.

@sgttris @amistre64 Ask yourself: would this alternating series be convergent or divergent, and why?

Answer 14

im working on the why part at the moment :)

Answer 15

my own mind goes this far
\[\sum_1\frac1{2n}+\sum_1\frac{-1}{2n-1}\]
\[\sum_1\frac1{2n}-\frac{1}{2n-1}\]
\[\sum_1\frac{2n-2}{2n(2n-1)}\]
\[\sum_1\frac{n-1}{n(2n-1)}\]

Answer 16

ugh, thats off for some reason

Answer 17

Please look up "alternating series" and refresh your memory regarding the criteria for convergence of an "alt series."

Answer 18

if lim to 0, and sequence is decreasing, then convergent seems to be the criteria

Answer 19

since limit of 1/n is 0, and 1/n is decreasing ... still want to see it a different way tho, i like things to be difficult

Answer 20

i also forgot how to add fraction ...
\[\sum_1\frac{2n-1-2n}{2n(2n-1)}\]
\[\sum_1\frac{-1}{2n(2n-1)}\]
thats better

Answer 21

We're trying to determine whether the series (-1)^n / n converges or diverges, and recognize this as an alternating series. Separate the (-1)^n from the rest, focusing on \[a _{n}=1/n\]

If: (1) this a-sub-n is decreasing with n
(2) its limit as n goes to infinity is zero
and
(3) it's always positive,
then we conclude tht the given alternating series converges.

Answer 22

hmm, the 3rd crit is new to me

Answer 23

Yes, I was "winging it." How about this: YOU look up "alternating series" and determine more accurately than I what the third criterion is? :)

Answer 24

yours resembles Leibnez thrm for alternating series. according to math24.net

Answer 25

i cant seem to find it on Pauls site ... which is usually pretty good about that stuff :/

Answer 26

Paul seems to be saying that as long as |an| eventually conforms, then its convergent .... shich seems to me to imply that it can be wonky at first but evens out in the end

Answer 27

idk how to type it and I'm in a rush but if A sub n+1 is less than or equal to A sub n and the lim of n to infinity is =0 then it will converge

Answer 28

*lim of n to infinity for A sub n

Answer 29

Yes, that was one of the criteria for convergence: the limit as n goes to inf. of a-sub-n must be zero. I see that both of you know enough about this material so that a quick ref to material on "alternating series" will give you all 3 criteria. You interested enough to check that out (gasp, in a book)? :)

Answer 30

my archie and jughead doesnt have a chapter on it ;)

Answer 31

hm, I was only taught that, and I wish I had a book lol, thank you though, the more I know the better

Answer 32

@amistre64: Don't depend upon Archie or Jughead. Both are numbskulls. Ask Veronica instead.

Answer 33

@sgttris: I surely appreciate your interest in this material.