Question

Calculus. Limits

Answer 1

\[a_n = 6n \cos (npi)\]

Answer 2

Determine whether the sequence converges or diverges. If it converges, find the limit

Answer 3

test?

Answer 4

oh

Answer 5

idk where to start

Answer 6

i was hoping to find out how to figure out whether its convergent or divergent and the limit if it is convergent

Answer 7

i know that cos oscillates between -1 and 1... but how do i test for convergence with that?

Answer 8

freckles!

Answer 9

\[\lim_{n \rightarrow \infty}a_n=\lim_{n \rightarrow \infty}6 n \cos( n \pi) \\ \]
So we know we have to look at this limit?

Answer 10

yes

Answer 11

is it a plug and chugg thing? where I can try different values for n?

Answer 12

\[-1 \le \cos( n \pi) \le 1 \]
as you said
now assuming if n>0 which we can do that since n-> positive infinity
the inequality is if we multiply by n:
\[-n \le n \cos( n \pi) \le n \]
I think this should show that your limit alternates between -inf and inf

Answer 13

oscillates*

Answer 14

yes. and what does that mean as far as converging and diverging goes?

Answer 15

it doesn't get close to any particular number
it doesn't even go to one particular infinity
the limit does not exist

Answer 16

oh wow~!

Answer 17

ok. so, anytime i have a function of cosine... that is set up with multiplication like this... does that always mean DNE?

Answer 18

just like you would say sin(infty) does not exist because this limit oscillates between -1 and 1

Answer 19

What about this one:
\[\lim_{n \rightarrow \infty}\frac{\sin(n)}{n} \text{ or } \lim_{n \rightarrow \infty}\frac{\cos(n)}{n}\]

Answer 20

On my school website, for this particular question, it did not give us an option to say DNE. It tells us to give a value for convergent or write none if it is divergent. I typed NONE for this and it said it was correct. Does this imply divergent and DNE are the same thing?

Answer 21

I would say that one is convergent at zero because 1/n= zero

Answer 22

if it isn't convergent then it is divergent
you are certainly correct to type none

Answer 23

Once again freckles, thank you!

Answer 24

yeah both of those I just mention converge to 0
since -1/n<sin(n)/n<1/n
and both -1/n and 1/n go to 0 as n goes to inf
so sin(n)/n->0 as n goes to inf

Answer 25

we could put cos there instead of sin

Answer 26

so, all of these rules are the same for both cos and sin?

Answer 27

I am going to close this question, so I can open a new one....

Answer 28

Well not exactly... \[\lim_{n \rightarrow \infty}n \cos(\frac{1}{n})= \infty \\ \ \text{ while } \\ \lim_{n \rightarrow \infty} n \sin(\frac{1}{n})=1\]

Answer 29

ohhh... I see!

Answer 30

\[\lim_{n \rightarrow \infty} n \cos(\frac{1}{n})=\lim_{u \rightarrow 0^+} \frac{1}{u} \cos(u)=\lim_{u \rightarrow 0^+}\frac{\cos(u)}{u} =\infty \\ \text{ but the other limit } \\ \lim_{n \rightarrow \infty}n \sin(\frac{1}{n})=\lim_{u \rightarrow 0^+} \frac{\sin(u)}{u}=1 \text{ is a well known limit }\]