How to find the $\lim\limits_{n \to\infty} \frac{2^{n^2}}{(n!)^2}$ 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. 
 A: $a_n=\frac{2^{n^2}}{(n!)^2}$
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 $a_n \to +\infty$
A: 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.
A: 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
$$\frac{u_{n+1}}{u_n}=\frac{2^{(n+1)^2}}{((n+1)!)^2}\cdot\frac{(n!)^2}{2^{n^2}}=
\frac{2^{(n+1)^2-n^2}}{(n+1)^2}=\frac{2^{2n+1}}{(n+1)^2}\to+\infty.$$
A: 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.
A: 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.
A: 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:
$$\lim_{n \rightarrow + \infty} \dfrac{2(n^2)}{(n!)^2} = \lim_{n \rightarrow + \infty} \dfrac{2(n^2)}{(n(n-1)!)^2} = \lim_{n \rightarrow + \infty} \dfrac{2(n^2)}{n^2((n-1)!)^2} = \lim_{n \rightarrow + \infty} \dfrac{2}{((n-1)!)^2} = 0.$$
