Question

How to prove the sequence {cos(n)} diverges

Answer 1

What tests do you have? Is cosine bounded? if so by what?

Answer 2

lcos(n)l <=1 , can it be done by sqeeze theorem ?

Answer 3

I wasn't thinking that, but actually, I think it can what function would you use?

Answer 4

i don't know, so i would like to know if there is another way :/

Answer 5

I think squeeze will be easiest

Answer 6

why don't we do harmonic?

Answer 7

like 2/n

Answer 8

value of cos n oscillates between -1 and 1,hence it do not converge.

Answer 9

need a full proof

Answer 10

though I agree, think they need something a bit more substantial

Answer 11

is is possible for a sequence to be bounded and convergent ? if you said so

Answer 12

? sorry can you rephrase that, i don't understand your question

Answer 13

Surjihayer proved it's divergent by saying its oscillations between -1 and 1, but i would like to know if it's possible for {cos(n)} to be bounded by -1 and 1, and still convergent. Clearly i just need proof

Answer 14

oh it's a thing for cosine, sine but ie 1/n^2 is convergent, but bounded

Answer 15

actually in order to converge, it has to be bounded i believe, but anyways let's try squeeze

Answer 16

so can you think of a function that diverges which is always greater than cosine?

Answer 17

i think the sqeeze theorem is usually compared with a definite number, if you have to think of another function, i think it's Comparison Test

Answer 18

meh, semantics, I think of them as one in the same you compare by picking a fn then their limits squeeze an answer

Answer 19

but my initial thought was the integral test, though I doubt you've proven sin's div if you haven't done cos

Answer 20

could just convert it into it's series form. basic but simple way

Answer 21

how can we convert a sequence into a series? I thought we only can covert a function into a series ?

Answer 22

meh true i was thinking in series, partial sums( subsequences)? ie n>0 n=0 n<0

Answer 23

n=0 yields 1+1+1 --> diverge

Answer 24

okay

Answer 25

check the others maybe

Answer 26

uhm hmm

Answer 27

i mean it contains a divergent subsequence, wouldn't that be enough?

Answer 28

n=2kpi not 0

Answer 29

i think it's enough to prove it's divergent, maybe i use that way by partial sums, thankyou

Answer 30

np sorry, I forget how we acually proved proved it it was like a common sense thing

Answer 31

but i mean if you graph it the tops cancel out the bottoms right?

Answer 32

so you are left with 1+1+1+1....

Answer 33

that's how i'd go about it without using any tests I think

Answer 34

maybe that's it, there's another way to prove it by epsilon, but i'm not quite sure by that way either

Answer 35

hmm i'd have to think on it

Answer 36

you mean by def of a limit right?

Answer 37

yeah, it's a hard way to do though

Answer 38

can it be of any help?
\[\cos x=1-\frac{ x^2 }{ 2! }+\frac{ x^4 }{ 4! }-\frac{ x^6 }{ 6! }+...+\frac{ \left( -1 \right)^kx ^{2k} }{ 2k! }+...\]

Answer 39

we had suggested that, but it's a sequence not a series, so that would have to be done for every term

Answer 40

which would be a little much unless you can think of a way to apply it

Answer 41

Can't we just do the divergence test? Since the limit as n approaches infinity of cos(n) is not 0, then we know it must diverge.

Answer 42

we'd have to prove that test works first

Answer 43

but yea, since there is no limit that would work

Answer 44

I totally forgot about that one