Question

Does the series tan 4/n converge or diverge?

Answer 1

diverge. \[\sum_{n=1}^{\infty} \tan(4/n) = \sum_{n=1}^{\infty} \sin(4/n)/\cos(4/n)\]
but we know that 0<=1sin(x)<=1 and cos 0<=cos <=1, so the limit doesn't =0, and hence fails according to the divergence test.

Answer 2

wait what, sin (0) is 0 and cos (0) = 1 so the limit of a_n is 0.

Answer 3

yes maybe, but the point is its a series. its an infinite amount of additions, and sin and cos oscillation between -1 and 1 (soz, it wasn't meant to be zero above, rather -1)

Answer 4

so, it can't coverage. it oscillates, so the limit doesn't even exist.

Answer 5

Can I use the comparison test? because I don't think that answer is enough for my final.

Answer 6

compare with what?

Answer 7

a smaller series so that if that smaller series diverges, the larger one diverges as well.

Answer 8

dont belive so

Answer 9

wolfram alpha says the comparison test, I just don't know how to justify it.

Answer 10

I mean it doesn't tell me the comparison.

Answer 11

acutally i think im wrong

Answer 12

I got it. The comparison is to 1/n. Just divide, use Lhopitals.

Answer 13

it does indeed diverge

Answer 14

or use the identity sin n/n as n->0 is 1.

Answer 15

ye true

Answer 16

My mistake was I kept trying to justify how 1/n was smaller than tan 1/n but all I needed to do was divide.

Answer 17

thanks for your help anyway.

Answer 18

I appreciate the attempt nonetheless.

Answer 19

hah np, im learning with you man. got my test in 1 month, so thanks for the question

Answer 20

nice, add me to your friends list. we can help each other. my test is on monday and my final is 2 weeks afterwards.

Answer 21

k, cool, tell me your answers lol/

Answer 22

anway, g2g cya

Answer 23

cya