In my exam there was this question :
Prove convergence or divergence of :
$\displaystyle \sum_{n=1}^ \infty \frac{(n!)^24^n}{(2n)!}$
I noticed that :
$\displaystyle \lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!}=\frac{4^n 2\pi n \left(\frac{n}{e} \right)^{2n}}{\left(\frac{2n}{e} \right)^{2n }\sqrt{2\pi (2n)}} = \lim_{n \to \infty } \sqrt{\pi n}= \infty$
I couldn't prove this without using Stirling's approximation
I tried to show that $\displaystyle\prod_{k=1}^n \frac{1+\frac{n}{k}}{4} \to 0 $ but that was not possible for me
I also tried to use Stolz cesaro theorem but I couldn't show that the limit $\rightarrow \infty$
(all I got was $\displaystyle \lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!}= \lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!} \times \frac{4(n+1^2)-1}{(2n+1)(2n+2)-1}=\lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!}$ )
Best Answer
First we split $(2n)!$ into product of even factors $\cdot$ product of odd factors i.e. : $(2n)! = (2n)(2n-2)..(2) \cdot (2n-1)(2n-3)..(1)$
Now we write $4^n = (2^2)^n = 2^{2n} = 2^n \cdot 2^n$
We multiply both $2^n$ to both $(n!)$ to get $[(2n)(2n-2)..(2)]^2 = (2n)(2n-2)..(2) \cdot (2n)(2n-2)..(2)$
So we have the last term $\lim_{n \rightarrow \infty} \frac{(n!)^2 \cdot 4^n}{(2n)!} = \lim_{n \rightarrow \infty} \frac{(2n)(2n-2)..(2) \cdot (2n)(2n-2)..(2)}{(2n)(2n-2)..(2) \cdot (2n-1)(2n-3)..(1)} = \lim_{n \rightarrow \infty} \frac{(2n)(2n-2)..(2)}{(2n-1)(2n-3)(2n-5)..(1)}$
Now we have $2n > 2n-1 \;\&\; 2n-2 > 2n-3 \;\&\; 2n-4 > 2n-5 .. 2 > 1$
$\therefore \lim_{n \rightarrow \infty} \frac{(2n)(2n-2)..(2)}{(2n-1)(2n-3)(2n-5)..(1)} = \lim_{n \rightarrow \infty} \frac{2n}{2n-1} \cdot \frac{2n-2}{2n-3} \cdot \frac{2n-4}{2n-5} ... \frac{2}{1}$
Since this is a product of infinite terms and all are greater than 1, the product $\rightarrow \infty$