How can I evaluate the limit . $$L=\lim _{n\rightarrow \infty} \frac{1}{n} \sum _{z=1}^{n} z!^{1/z^2 }=\lim _{n\rightarrow \infty} \frac{1+\sqrt[2^2]{2!}+\sqrt[3^2]{3!}+\cdots+\sqrt[n^2]{n!}}{n} $$

  • $\begingroup$ What have you tried so far? $\endgroup$ – Klangen Sep 11 at 10:18
  • $\begingroup$ The wolfram alpha shows the answer is complex infinity $\endgroup$ – Isaac YIU Math Studio Sep 11 at 10:22
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    $\begingroup$ Wolfram alpha is dumb. The limit is obviously 1, because the summands in the numerator tend to 1 and there are $n$ of them. $\endgroup$ – Ivan Neretin Sep 11 at 10:29
  • $\begingroup$ I agree with Ivan. $\lim_{z \to +\infty} z!^{1/z^2} = 1$, so these Cesaro means also converge to $1$. $\endgroup$ – GEdgar Sep 11 at 10:37
  • $\begingroup$ $n!=1\cdot 2\cdots n\leq n \cdot n \cdots n = n^n$ then easily $(n!)^{1/n^2}\leq n^{1/n}\to 1$ $\endgroup$ – kingW3 Sep 11 at 11:38

Using Stolz-Cesaro you need only calculate the limit

  • $\lim_{n\to \infty}(n!)^{\frac{1}{n^2}}$

You have because of GM-AM $$1 \leq (n!)^{\frac{1}{n^2}} \stackrel{GM-AM}{\leq} \left(\frac{\frac{n(n+1)}{2}}{n}\right)^{\frac{1}{n}} = \sqrt[n]{\frac{n+1}{2}}\stackrel{n \to \infty}{\longrightarrow}1$$


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