What is the $\lim_{n\to\infty}\frac{e^{n^{2}}}{(2n)!}$?

Question

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.

Answer 1

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.

Answer 2

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 $\infty$).

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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?

Answer 3

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 $n \rightarrow +\infty$.