Simplify and find $\lim_{n\to \infty}\frac{(2n-1)!}{(2n+1)!}$

Question

So I was calculating $$\lim_{n\to \infty}\frac{(2n-1)!}{(2n+1)!}$$ and couldn't solve it, so I saw the answer sheet and it said that the limit was $0$, I checked the process and they simplified the expression to $$\lim_{n\to\infty} \frac{(2n-1)!}{(2n + 1)(2n)(2n − 1)!}$$ and then $$\frac{1}{(2n + 1)(2n)}$$ and said it was $0$

How did they achieved this?

Answer 1

Let's take a smaller example. Consider $$5! = 5\cdot4\cdot3\cdot2\cdot1.$$ Notice that this can be expressed as $$5! = 5\cdot(4\cdot3\cdot2\cdot1\cdot) = 5\cdot4!.$$ Similarly $$5! = 5\cdot4\cdot(3\cdot2\cdot1) = 5\cdot4\cdot3!$$ We can generalize to $k!$, $$k! = k(k-1)(k-2)\dotsm 2\cdot1$$ and in particular $$k! = k(k-1)(k-2)!\tag a$$

Now, \begin{align*} \lim_{n\to\infty} \frac{(2n-1)!}{(2n+1)!} &= \lim_{n\to\infty}\frac{(2n-1)!}{(2n+1)(2n)(2n-1)!}\tag b\\ &=\lim_{n\to\infty}\frac{1}{(2n+1)(2n)}\\ &= 0 \end{align*} where in (b), I could use (a) as an aid by letting $k = 2n+1$.

Answer 2

Hint: note that if the series $$\sum_{n=1}^{+ \infty}\frac{(2n-1)!}{(2n+1)!}$$ converges $$\lim_{n\to \infty}\frac{(2n-1)!}{(2n+1)!}=0$$ you can use the ratio test to show that the series converges.