Thread: When taking the limit of an infinite series...

When you cos[n(x+1)] as n-> , at one point you say lim n-> 1/n [sin2n -sin(n)] lim n-> 2/n = 0 .

Why is it okay to say here? How do you know that the integral is 0, and not something less than 0.

Originally Posted by gummy_ratz

When you integrate cos[n(x+1)] as n->00 from 0 to 1, at one point you say lim n->00 1/n[sin2n -sin(n)] <= lim n->oo 2/n = 0.

Why is it okay to say <= here? How do you know that the integral is 0, and not something less than 0.

Hi gummy_ratx,

It is not clear what your problem is. Can you use a bit of latex to write your expression. http://www.mathhelpforum.com/math-he...orial-266.html