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December 17th, 2017, 12:55 PM   #1
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Evaluate limit

$\displaystyle \lim _{n\rightarrow \infty} \frac{1}{n} \sum _{z=1}^{n} z!^{\frac{1}{z^2 }}=\lim _{n\rightarrow \infty} \frac{1+\sqrt[2^2]{2!}+\sqrt[3^2]{3!}+...+\sqrt[n^2]{n!}}{n}$
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December 19th, 2017, 09:29 AM   #2
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My solution
Using Cesaro-Stolz theory : $\displaystyle \sqrt[(n+1)^{2}]{(n+1)!}\rightarrow \frac{1}{n} \sum _{z=1}^{n} z!^{\frac{1}{z^2 }} $ gives $\displaystyle \lim_{n\rightarrow \infty} \sqrt[(n+1)^{2}]{(n+1)!}=\lim_{n\rightarrow \infty} \frac{1}{n} \sum _{z=1}^{n} z!^{\frac{1}{z^2 }}$
Using AM-GM : $\displaystyle 1\leq \sqrt[(n+1)^{2}]{(n+1)!}<\sqrt[n+1]{\frac{n+2}{2}} $
$\displaystyle 1\leq \lim_{n\rightarrow \infty} \sqrt[(n+1)^{2}]{(n+1)!}<\lim_{n\rightarrow \infty} \sqrt[n+1]{\frac{n+2}{2}}=1 $ so $\displaystyle \; 1\leq L<1 \;$ gives $\displaystyle L=1$
$\displaystyle \lim_{n\rightarrow \infty} \frac{1}{n} \sum _{z=1}^{n} z!^{\frac{1}{z^2 }}=1$
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