How to find the $\lim\limits_{n \to\infty} \frac{2^{n^2}}{(n!)^2}$

Question

I tried to compute the limit above and got stuck. I thought using L'Hospital's Rule would be a good idea, I got : $\lim\limits_{n \to\infty} \frac{\exp(n^2\ln(2))}{(n!)^2}$ But I found out that it wasn't so useful to me because I don't know how to derivate a factorial and what is in the exponential cannot be simplified with Ll'Hospital.

Does anybody have a clue of what I could do?

Thanks in advance.

Answer 1

First of all, I suppose that the limit you want to compute is the following $$\lim_{n \rightarrow + \infty} \dfrac{2(n^2)}{(n!)^2}.$$

You can proceed like this: $$\lim_{n \rightarrow + \infty} \dfrac{2(n^2)}{(n!)^2} = \lim_{n \rightarrow + \infty} \dfrac{2(n^2)}{(n(n-1)!)^2} = \lim_{n \rightarrow + \infty} \dfrac{2(n^2)}{n^2((n-1)!)^2} = \lim_{n \rightarrow + \infty} \dfrac{2}{((n-1)!)^2} = 0.$$

Answer 2

$a_n=\frac{2^{n^2}}{(n!)^2}$

We can easily show that $\frac{a_{n+1}}{a_n} \to +\infty$

Thus exists $m \in \Bbb{N}$ such that $\frac{a_{n+1}}{a_n}>2,\forall n\geq m$

From this you can prove that $a_n \to +\infty$

Answer 3

If the sequence $u_n=\dfrac{2^{n^2}}{(n!)^2}$ converges to a (finite) limit, two consecutive terms must be equivalent, {2^{n^2}}{(n!)^2}i.e. $\;\lim_{n\to\infty}\dfrac{u_{n+1}}{u_n}=1$. Now $$\frac{u_{n+1}}{u_n}=\frac{2^{(n+1)^2}}{((n+1)!)^2}\cdot\frac{(n!)^2}{2^{n^2}}= \frac{2^{(n+1)^2-n^2}}{(n+1)^2}=\frac{2^{2n+1}}{(n+1)^2}\to+\infty.$$

Answer 4

Hint:

If you increase $n$ by one, the numerator gets mutliplied by $2^{2n+1}$ while the denominator gets multiplied by $(n+1)^2$. The numerator is a clear winner.

Answer 5

You could state that: $$ u_n = \frac{2^{\left(n^2\right)}}{n!^2}$$ find that: $$\forall n \in \mathbb{N}, u_{n+1}= \frac{2^{(n+1)^2}}{(n+1)!^2}= \frac{2^{n^2+2n+1}}{((n+1)n!)^2}=\frac{2^{n^2}}{n!^2} \frac{2^{2n+1}}{(n+1)^2}=u_n \times \frac{2^{2n+1}}{(n+1)^2}$$ and see that (you can easily prove it by induction): $$ \forall n \in \mathbb{N}, a_n= \frac{2^{2n+1}}{(n+1)^2} \gt 2 $$ Now, you have: $$\forall n \in \mathbb{N^*}, u_n = u_0\prod\limits_{i=1}^na_i \gt u_0 \times 2^n$$ Because $2^n$ diverges, $u_n$ does too.

Answer 6

Notice $$n!\le n^n\implies \frac{1}{(n!)^2}\ge\frac{1}{n^{2n}}=e^{-2n\ln n}$$

And

$$\lim_{n\to\infty}\frac{\ln n}{n}\to 0$$

So the limit

$$L\ge\lim_{n\to\infty}2^{n^2}e^{-2n\ln n}=\lim_{n\to\infty}e^{n^2\ln 2-2n\ln n}=\lim_{n\to\infty}e^{2n^2(\frac{\ln2}{2}-\frac{\ln n}{n})}\to\infty$$

diverges.