How to calculate $\lim_{k \to \infty} \frac{(2k)!}{(2k+2)!}$

Question

I need to calculate limit number 1, and I don't understand how to get out the factors.

$$ (1) \lim_{k \to \infty} \frac{(2k)!}{(2k+2)!}$$

$$ (2) \lim_{k \to \infty} \frac{(k)!}{(k+1)!}$$

When I calculate limit number $2$, I cancel the factors then I get :

$$ (2) \lim_{k \to \infty} \frac{(k)!}{(k+1)!}=\lim_{k \to \infty}\frac{1}{k+1}=0$$

Using the same method use in limit $(1)$ When I calculate limit number $1$, I cancel the factors then I get :

$$ (1) \lim_{k \to \infty} \frac{(2k)!}{(2k+2)!} = \lim_{k \to \infty} \frac{2}{2k+2}$$

But this is inerror, I don't understand how I should get out the factors from limit $(2)$.

Any help will be appreciated.

Answer 1

Hint: For $(1)$, we have $(2k+2)!=(2k+2)(2k+1)(2k)!$, which implies that $$\frac{(2k)!}{(2k+2)!}=\frac{(2k)!}{(2k+2)(2k+1)(2k)!}=\frac{1}{(2k+2)(2k+1)}.$$

Answer 2

hint: $(k+1)! = (k+1)\cdot k!$

Answer 3

HINT: $(2k+2)!=(2k+2)(2k+1)(2k)!$