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

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

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

it goes to 0 that makes it inconclusive

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...

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

oh wait, maybe it will work!

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

i think they are the same

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

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

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

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

l'hopital gets it in one step i believe

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

whew!

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

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

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

@satellite73

Limit comparison test, not ratio test.

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