Question

does sum (1 to inf) of tan (1/n) converges or diverges?

Answer 1

Oh, sorry, I'm getting ahead of myself. That test doesn't help...

Answer 2

Yes, I know. I was planning on using the test for divergence, which involves a limit. But it doesn't work for this series.

Answer 3

it goes to 0 that makes it inconclusive

Answer 4

I'm thinking comparison test. For sufficiently large n, you have \(\large\tan\left(\frac{1}{n}\right)\le\frac{1}{n}.\) Let me verify that...

Answer 5

hmm i tried that but it didn't seem to work
or rather i tries \(\tan(\frac{1}{n})>\frac{1}{n}\) to show it diverges, but that does not work

Answer 6

oh wait, maybe it will work!

Answer 7

What about \(\large\frac{2}{n}\)?

Answer 8

i think they are the same

Answer 9

constant does not play role in comparison test generally so its same

Answer 10

i mean i think
\[\tan(\frac{1}{n})\sim\frac{1}{n}\]

Answer 11

i think
can check that
\[\lim_{n\to \infty}\frac{\tan(\frac{1}{n})}{\frac{1}{n}}=1\]

Answer 12

Yeah, I was just checking that myself. It's correct.

Answer 13

l'hopital gets it in one step i believe

Answer 14

And wolfram also says that the series diverges by comparison test: http://www.wolframalpha.com/input/?i=series+tan%281%2Fn%29+from+n%3D1+to+infinity

Answer 15

whew!

Answer 16

try l'hopital, it will give the limit as 1, so diverges by comparison

Answer 17

guys, in that case limit = 1, when limit = 1, test is inconclusive again

Answer 18

in ratio test, if limit < 1, it converge, limit > 1 , it diverge and limit= 1 , its inconclusive

Answer 19

@satellite73

Answer 20

Limit comparison test, not ratio test.