1. ## series limit RP

$lim_{n \to \infty} \frac{1*3*5*....*(2n-1)}{2*4*6*...*(2n)} = ?$

2. Mistaken post. I can't remove it.

3. ?

4. (2)(4)(6)...(2n-2)(2n)= (2*1)(2*2)(2*3)...(2*(n-1))(2*n)= $2^n n!$.

1(3)(5)...(2n-1)= $\frac{1(2)(3)(4)...(2n-1)(2n)}{2(4)(6)...(2n)}= \frac{(2n)!}{2^n n!}$ so

$\frac{1*3*5*\cdot\cdot\cdot(2n-1)}{2*4*6*\cdot\cdot\cdot (2n)}= \frac{(2n)!}{2^{2n}n!}$

That $2^{2n}$ in the denominator "dominates" is enough to show that the limit is 0.

5. Originally Posted by HallsofIvy
(2)(4)(6)...(2n-2)(2n)= (2*1)(2*2)(2*3)...(2*(n-1))(2*n)= $2^n n!$.

1(3)(5)...(2n-1)= $\frac{1(2)(3)(4)...(2n-1)(2n)}{2(4)(6)...(2n)}= \frac{(2n)!}{2^n n!}$ so

$\frac{1*3*5*\cdot\cdot\cdot(2n-1)}{2*4*6*\cdot\cdot\cdot (2n)}= \frac{(2n)!}{2^{2n}n!}$

That $2^{2n}$ in the denominator "dominates" is enough to show that the limit is 0.
Dear HallsofIvy,

I think you have made a slight error. The last line must be,
$\frac{1*3*5*\cdot\cdot\cdot(2n-1)}{2*4*6*\cdot\cdot\cdot (2n)}= \frac{(2n)!}{2^{2n}(n!)^2}$