Recent content by zak1989

Oh ok so the limit as n goes to infinity is zero which means it converges, but I'm a little lost. How do I know the limit is zero? Thanks for all your help I really appreciate it :)

Ok now I understand why those two equations are equal, thanks. It seems so obvious now. What I still don't understand is how does saying 1/2n * 3/2n * 5/2n...(2n-1)/2n less than 1/2n help me conclude that the sequence diverges?

Thanks for your help, but I'm still not getting it. When I see the sequence an = (1*3*5*...*(2n-1))/(2n)^n I would think it looks something like 1/2(1)^1 * 3/2(2)^2 * 5/2(3)^3 ... Is that wrong? I don't understand where the exponent n is going from the denominator.

Homework Statement Determine the convergence or divergence of the sequence with the given nth term. If the sequence converges find its limit. an = (1*3*5*...*(2n-1))/(2n)n Homework Equations lim n->infinity an = L The Attempt at a Solution The answer in the book shows: 1/2n *...

Thanks for the help guys, but how would I go about doing it without partial fractions? I'm not supposed to use that technique yet. Sorry I forgot to mention that in my post.

Homework Statement Ok the problem is: ∫-1/(4x-x^2) dx The answer in the back of the book is: (1/4)ln(abs((x-4)/x))) + C Homework Equations I think this would be used somehow: ∫ du/(a^2-u^2) = 1/2a ln(abs((a+u)/(a-u))) + C The Attempt at a Solution ∫-1/(4x-x^2) dx...