What is the $\lim_{n\to\infty}\frac{e^{n^{2}}}{(2n)!}$? Evaluate $\lim_{n\to\infty}\frac{e^{n^{2}}}{(2n)!}$.
I used the ratio test to calculate this limit, and I got here:
$\lim_{n\to\infty}\frac{e^{2n}}{n^2}\frac{e}{4+6/n+2/n^2}$.
For the first fraction, I applied the ratio test again and got +inf.
I also attach pictures with a more detailed solution. I would like to ask if the procedure and result are correct.


 A: perhaps it becomes clearer that the limit is infinite, making the series expansion of the exponential
$$   \dfrac{e^{n^2}}{(2n)!} = \dfrac{\sum_{m=0}^{\infty}\dfrac{(n^2)^m}{m!} }{(2n)!} \geq  \dfrac{n^{4n}}{  
 ( (2n)!)^2 }=   \left(  \dfrac{n^{2n}}{  
  (2n)! } \right)^2 $$
the last expression on the right is $\geq Cn$ ($C>0$) for large n.
A: HINT:
Note that we can write $\displaystyle \frac{e^{n^2}}{(2n)!}$ as
$$\frac{e^{n^2}}{(2n)!}=e^{n\left(n-\frac1n\log((2n)!)\right)}$$
Now, show that $\lim_{n\to\infty}\left(n-\frac1n\log((2n)!)\right)>0$ (in fact, the limit is $\infty$).


HINT $2$:
Use the fact that
$$\begin{align}
\frac1n\log((2n)!&=2\log(2n)+\frac2{2n}\sum_{k=1}^{2n}\log(k/2n)\\\\
&=2\log(2n)-2 +o(1)\,\,\text{as}\,\,n\to\infty
\end{align}$$
Can you proceed now?
A: By Stirling's approximation
$$\frac{e^{n^{2}}}{(2n)!} \sim \frac{e^{n^2} e^{2n}}{\sqrt{4\pi n} (2n)^{2n}} \sim \frac{1}{\sqrt{4\pi }} \exp \left(n^2 + 2n - 2n \log(2n) - \frac{1}{2}\log(n) \right) \longrightarrow +\infty$$
as $n \rightarrow +\infty$.
