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.

enter image description here

enter image description here


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.


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 $n \rightarrow +\infty$.