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Question

Deje $a_n = (n!)^{\frac{1}{n^2}}$ .

Ahora, $$n! \geq1 \implies (n!)^{\frac{1}{n^2}} \geq 1$$ and $$n! \leq n^n \implies (n!)^{\frac{1}{n^2}} \leq n^{\frac1n}$ $

Pero $$\lim_{n \to \infty} n^{\frac1n} = 1 = \lim_{n \to \infty} 1$$ Thus by Sandwich Theorem $$\lim_{n \to \infty} (n!)^{\frac{1}{n^2}} =1$ $ ¿Es correcta la solución?

Answer 1

Alternativamente: $$ \ lim_ {n \ to \ infty} \ log (n!) ^ {\ Frac {1} {n ^ 2}} = \ lim_ {n \ to \ infty} \ frac {\ log (n! )} {n ^ 2} = \ lim_ {n \ to \ infty} \ frac {\ log 1 + \ log 2 + \ cdots + \ log (n)} {n ^ 2} = $$ $$ \ lim_ { n \ to \ infty} \ frac {\ log (n + 1)} {(n + 1) ^ 2 - n ^ 2} = \ lim_ {n \ to \ infty} \ frac {\ log (n + 1) } {2n + 1} = 0. $$ (Stolz-Cesàro usado en el 3er $=$ )