Series from 1 to infinity of

sin(1/n)
n^(2/3)

I know I'm supposed to use the comparison test but can't find another series to use. 1/n^(2/3) is divergent so that doesn't help...

This serie converges.

Toward infinity, 1/n ~ 0 then sin(1/n) ~ 1/n.

Hence,

sin(1/n) ~ 1
n^(2/3) n^(5/3)


and 1/n^(5/3) have a convergent serie.

What are you exactly asked for ?

This post was edited by feanur on Mar 15 2014 05:39pm

'feanur' gave you the right answer, just would like to add that from the taylor series of sine you can see that sin(1/n) < 1/n so you can replace the "~" with a "<"
if you have to find what the series converges to you will in my opinion have to use the taylor series and fiddle around