Question

Use Cauchy criterion to tell if series converges or not

Answer 1

\[\frac{ \cos x }{ 2 }+\frac{ \cos 2x }{ 2^{2} }+....+\frac{ \cos nx }{ 2^{n} }+...\]

Answer 2

does this series converge or not?

Answer 3

@kropot72 @zepdrix
any ideas?

Answer 4

@zzr0ck3r can you help?

Answer 5

@satellite73

Answer 6

heeeellllppppppp please :-) :-)

Answer 7

@UnkleRhaukus can you help by chance?

Answer 8

i am going to say "yes" and then look up "cauchy"
in any case each of these terms is bounded by \(\frac{1}{2^n}\)

Answer 9

how do i use the cauchy criterion though to find out? u saying just b/c each term is bounded by 1/2^n

Answer 10

given any \(\epsilon>0\) there is an \(N\) such that for all \(n,m>N\) you have \(|a_n-a_m|<\epsilon\)
numerator is bounded above by \(1\) and below by \(-1\) so largest this can be is
\[\frac{1}{2^n}+\frac{1}{2^m}\]

Answer 11

at least i think that is right, maybe get a second opinion

Answer 12

i have been trying for awhile as you can see, haha., you are the first to respond though, im not sure i will get lucky enough to get another response...

does that say 1/2^n + 1/2^m
its tough to read on my computer.

Answer 13

\[\large \frac{1}{2^n}+\frac{1}{2^m}\]

Answer 14

is the largest \[\large |\frac{\cos(nx)}{2^n}-\frac{\cos(mx)}{2^m}|\] can be

Answer 15

so pick your \(N\) so that if \[\large n, m>N\] you get \[\large \frac{1}{2^n}+\frac{1}{2^m}<\epsilon\]

Answer 16

and if there is a sup, or max then it has to converge?

Answer 17

not sure what you mean
you have to show what i wrote above

Answer 18

check out this attachments

Answer 19

ok i might be wrong, let me think for a moment or two
you are trying to show that the series converges, not a sequence
the sequence would be the partial sums

Answer 20

yes, trying to show if the series converges or not.

Answer 21

yeah i was totally wrong, but the idea is right

Answer 22

replace \(a_n\) by \(S_n\) and you need to show that if for any \(\epsilon>0\) there is an \(N\) so that \(|S_n-S_m|<\epsilon\) which is the same as showing
\[\large| a_{n+1}+a_{n+2}+a_{n+3}+...+a_{n+p}|<\epsilon\]

Answer 23

in your case you have to show that
\[ \frac{\cos((n+1)x)}{2^{n+1}}+\frac{\cos((n+2)x)}{2^{n+2}}+\frac{\cos((n+13)x)}{2^{n+3}}+...+\frac{\cos((n+p)x)}{2^{n+p}}\] can be made arbitrarily small

Answer 24

but the numerator is still bounded above by 1, so the most this can be is
\[\frac{1}{2^{n+1}}+\frac{1}{2^{n+2}}+\frac{1}{2^{n+3}}+...+\frac{1}{2^{n+p}}\]

Answer 25

and this is less than or equal to
\[\frac{1}{2^{n+1}}+\frac{1}{2^{n+1}}+\frac{1}{2^{n+1}}+...\frac{1}{2^{n+1}}\] \(p\) times or
\[\frac{p}{2^{n+1}}\]

Answer 26

which you can supply the details to make this arbitrarily small, remembering that you get to pick \(N\) so that \(n>N\)

Answer 27

i appreciate you putting in the time on this one, this stuff is tough to understand for me, thank you for your time.