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...