Question

Determine whether the following sequence converge or diverge. Find the limit if it converges.

an= cos(2pi n)/n.

Okay, so seeing that this sequence is something big over small, I initially guessed that it diverges. But the answer is the limit goes to 0 and converges. I really don't see that. Shouldn't the sequence go to cos 2pi, since the n's cancel out on the top and bottom?

Answer 1

\[\frac{cos(2pi*n)}{n}?\]

Answer 2

yes

Answer 3

cos is the name of a function; the denominator doesnt work its way thru that to get to the other n

Answer 4

cos(2pi) = 1

and all multiples of it look to be 1 as well; this gives us the sequence related to:

1/n which diverges

Answer 5

the limit as n goes to 0?

Answer 6

0

Answer 7

so what happened to the n on top?

Answer 8

you use a bit of common sense to it :)

cos(2pi) = 1; and it doesnt matter how many times you hit that, your still at 1

Answer 9

cos(2pi*1) = 1
cos(2pi*2) = 1
cos(2pi*3) = 1
cos(2pi*4) = 1
cos(2pi*5) = 1
cos(2pi*n) = 1

Answer 10

oh, I see! thanks-you! :)

Answer 11

so this acts like: 1/n, which is the harmonic series

Answer 12

sequence or series?
sequence converges to zero for sure.
series (the sum) is divergent.

Answer 13

.... yeah, that would be good to keep track of, seq or ser ;)

Answer 14

so generally, something big/small diverges and small/big converges. can that always be reliable?

Answer 15

depends on what your defining as converge/diverge

Answer 16

\[a_n=\frac{1}{n}\]
\[\lim_{n->inf} a_n=0\]
the sequence itself converges

Answer 17

\[a_n=\frac{1}{n}\]
\[\sum_{n=1}^{inf} a_n=inf\]
the series diverges