Evaluate $\lim\limits_{n \to \infty} \sqrt[n^2]{1! + 2! + 3! + \dots + n!}$ $$\lim_{n \to \infty} \sqrt[n^2]{1! + 2! + 3! + \dots + n!}$$
I can try at least to evaluate it from bottom:
$$1 < \sqrt[n^2]{n^2} < \sqrt[n^2]{n!} < \sqrt[n^2]{1! + 2! + \dots + n!}$$
Well, it doesn't say anything. Ok, gonna try evaluation from top.
$$\sqrt[n^2]{1! + 2! + \cdots + n!} < \sqrt[n^2]{n \times n!}$$
More than that,
$$\lim_{n \to \infty} \frac{n!}{1! + 2! + \cdots + n!} = 1$$
but this equation is always less than one.  So instead of writing $\sqrt[n^2]{n \times n!}$ I can try fixed coefficient $1 < a < n$. It will work for $a = 2$, $a =1.5$, but I'm not sure though about $a = (1 + \frac{1}{n})$
$$\sqrt[n^2]{1! + 2! + \cdots + n!} < \sqrt[n^2]{a \times n!}$$
Just to get broad idea how it looks I'll assume that $a = 1$ and I'll try to use Stirling's approximation
$$\sqrt[n^2]{1 \times n!} \approx \left((2\pi n)^{\frac{1}{2}}\left(\frac{n}{e}\right)^n \right)^\frac{1}{n^2} = (2 \pi n)^{\frac{1}{2n^2}} \times \left(\frac{n}{e}\right)^{\frac{1}{n}}$$
$\lim_{n \to \infty} \sqrt[n]{n} = 1$, $2n^2 > 2 \pi n$, $\frac{n}{e} < n$ (for big numbers) mean, that
$$\lim_{n \to \infty}(2 \pi n)^{\frac{1}{2n^2}} \times (\frac{n}{e})^{\frac{1}{n}} \approx 1$$
So I can guess, that
$$\lim_{n \to \infty} \sqrt[n^2]{1! + 2! + 3! + \cdots + n!}  = 1$$
But I sincerely doubt that what I wrote here is barely a solution. Would you mind helping me? Hope I didn't make huge mistakes

Notice that :
$$1 \leq \sqrt[n^2]{1! + \cdots + n!} \leq \sqrt[n^2]{n! + \cdots + n!} = \sqrt[n^2]{n \, n!} \leq \sqrt[n^2]{n n^n} = \sqrt[n^2]{n^{n + 1}} \to 1$$
because :
$$\ln \sqrt[n^2]{n^{n + 1}} = \dfrac{1}{n^2} \ln n^{n + 1} = \dfrac{n + 1}{n} \, \dfrac{\ln n}{n} \to 1 \times 0 = 0$$
