Finding the limit of this factorial sequence as n tends to infinity

Question

Find the limit of this factorial sequence as $n$ tends to infinity

$$a_n= \frac{((n+1)!)((n-1)!)}{(2n)!}$$

What I tried so far:

I simplified above to $$\frac{n(n+1)((n-1)!)^2}{(2n)(2n-1)(2n-2)\cdots} = \frac{(n+1)((n-1)!)}{(2)2^{0.5n}(2n-1)(2n-3)(2n-5)\cdots}$$

and I dont know where to go from here or if this is even the right path

Answer 1

Hint: Apply the ratio test for the convergence of $\sum a_n$.

Solution:

$\frac{a_{n+1}}{a_n} \to \frac14$ and so $\sum a_n$ converges. This implies that $a_n \to 0$.

Answer 2

You can observe that $$2^n (n!)^2<(2n)! < 2^{2n} (n!)^2 $$

Then apply sandwich theorem on $a_n$.

$$\frac{(n+1)!(n-1)!}{2^{2n} (n!)^2} \le a_n\le \frac{(n+1)!(n-1)!}{2^n (n!)^2}$$

Apply $\lim_{n \to \infty}$. We see the limit comes out to be $0$.

Answer 3

Take logarithms of both sides $$a_n= \frac{((n+1)!)((n-1)!)}{(2n)!}\implies \log(a_n)=\log((n+1)!)+\log((n-1)!)-\log((2n)!)$$ and use Stirling approximation for $\log(p!)$.

You should end with $$\log(a_n)\sim -2 n \log (2)$$

Answer 4

we have theorem - if $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = l , |l| < 1$ then $a_n \to 0$

Usingthis we have $\lim_{n \to \infty}\frac{a_{n+1}}{a_n} = \frac{\frac {(n+2)! n!}{(2n+2)!}}{\frac{(n+1)!(n-1)!}{(2n)!}} = \frac{(1+\frac{2}{n})}{(2+\frac{1}{n})(2+\frac{2}{n})} = \frac{1}{4} < 1 \implies a_n \to 0$