Question

an = cos(npi/2). If convergent find limit of sequence?

Answer 1

an n goes to infinity?

Answer 2

as*

Answer 3

Determine whether the sequence is convergent or divergent. If it is convergent, find its limit.
(a) an = cos(npi/2)
This was the question.

Answer 4

so the sequence is:
\[\cos(\frac{\pi}{2}), \cos(\pi), \cos(\frac{3\pi}{2}), \cos(2\pi), \cos(\frac{5\pi}{2}), \ldots\]
Do konw the value of those cos expressions?

Answer 5

ugh >.< do you know* the value of those cos expressions.

Answer 6

No dude. I have a midterm tomorrow and I have no Idea on how do solve this type of questions. I posted the exact question that was in the review sheet. It is tested on sequences from calc 2.

Answer 7

Will you have access to a calculator during the test? its something you can type right in. If not, you will have to have your trig values memorized.

Answer 8

not really

Answer 9

you dont need to know what the values are, only understand how the cos function behaves

Answer 10

Your help is much appreciated. We can use a scientific calculator but not a graphing one.

Answer 11

well, I take that back. in second term calc, you really should know what those values are

Answer 12

1,0,-1,0,1,0,-1, etc to infinity

Answer 13

so it does not converge

Answer 14

I get it now. Since cos (npi/2) oscillates between -/+ 1 as n tends to infinity. There is no unique value. Therefore limit does not exist and the sequence is divergent.

Am I right?

Answer 15

Yes

Answer 16

Thank you Sir. [Both of you]

Answer 17

\[\forall n \in \mathbb{N} , -1\le \cos(\frac{n \pi}{2})\le1 \]
So the sequence is always contained in the interval [0;1]
But,
\[\lim_{n \rightarrow +\infty} 1 =1 \]
\[\lim_{n \rightarrow +\infty} -1 =-1 \]
So by using the squeeze theorem one concludes that the sequence doesn't converge.

That would be the rigurous way to answer it.

Answer 18

What about an = ln(n)/sqrt(n). Does it converge or diverge?

Answer 19

Use L'Hopital's Rule to simplify the problem.

Answer 20

for ^^ use the p rule if 1/n^p if p is greater than one it will diverge and if it is less than one it will converge it will look like this 1/n^1/p

to answer your original question, i dont know that one but i know sin and cos dont converge

Answer 21

They don't, but on paper you have to prove it.