how to prove that $\displaystyle \lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!}= \infty$ without using Stirling's approximation

Question

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 $\displaystyle \lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!}= \lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!} \times \frac{4(n+1^2)-1}{(2n+1)(2n+2)-1}=\lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!}$ )

Answer 1

Use the Quotient-Criterion in a bit more specific form.

With $a_n = \frac{(n!)^2 4^n}{(2n)!}$ you get $$\frac{a_{n+1}}{a_n}= 2\frac{n+1}{2n+1}=1+\frac 1{2n+1}$$

Hence, with $a_1=2$ we get using a telescoping product: $$a_{n+1} = 2\prod_{i=1}^n\left(1+\frac 1{2i+1}\right)$$

Now, we apply the standard criterion for infinite products:

Since $\sum_{i=1}^\infty\frac 1{2i+1} $ is divergent, so is the corresponding infinite product.

Answer 2

Hint: Use $(2n)!=(2n)!!(2n-1)!!<(2n)!!^2$.

Answer 3

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 $\rightarrow \infty$

Answer 4

We have $$4^n=2^{2n}>{2n\choose n}={(2n)!\over ( n!)^2}$$ Hence $$a_n={4^n(n!)^2\over (2n)!}\ge 1$$ so the series $\sum a_n$ cannot be convergent. Moreover the sequence $a_n$ tends to infinity at the rate $\sqrt{n}.$ Indeed $${a_n\over a_{n-1}}={2n\over 2n-1}$$ For $n\ge 2$ we have $$\sqrt{2n\over 2n-1}\sqrt{2n+1\over 2n}\le {2n\over 2n-1}\le \sqrt{2n\over 2n-1}\sqrt{2n-1\over 2n-2}$$ Thus $$\sqrt{2n+1\over 2n-1}\le {a_n\over a_{n-1}}\le \sqrt{n\over n-1} $$ As $a_1=2$ the latter implies $$2\sqrt{2n+1}\le a_n\le 2\sqrt{n}$$

Answer 5

$$L(n) :=\frac{(n!)^24^n}{(2n)!}$$ $$\sqrt[n]{L(n)}=\sqrt[n]{\frac{(n!)^24^n}{(2n)!}}=4 \sqrt[n]{\prod\limits_{k=1}^n\frac{k}{k+n}}$$

By HM-GM inequality

$$\frac{4n}{n-\sum\limits_{k=1}^n \frac{1}{1+\frac{k}{n}}} \le \sqrt[n]{L(n)} $$ $$\frac{4}{1-\frac{1}{n}\sum\limits_{k=1}^n \frac{1}{1+\frac{k}{n}}} \le \sqrt[n]{L(n)} $$

$$\lim_{n\to \infty }\sqrt[n]{L(n)} \ge \frac{4}{1-\int_0^1\frac{1}{1+x}dx}> 3$$

since $\lim_{n\to \infty }\sqrt[n]{L(n)} >1$ there exist $N \in\mathbb{N}$ st $\sqrt[n]{L(n)} >2$ for any $n \ge N$

so $\sqrt[n]{L(n)}>2$ so $L(n)>2^{n}$ letting $n \to \infty $ we obtain:

$$\lim_{n\to \infty } L(n) =\infty $$