I know that $\lim_{n\to\infty} (n!)^{1/n} = \infty$ But I can't figure out the $\lim_{n\to\infty} [(1/n)(n!)^{1/n}]$ since it is infinity over zero. Please help.
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See N. S.'s answer here. – David Mitra Oct 07 '15 at 10:07
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You can also compute $\lim {a_{n+1}\over a_n}$ with $a_n=n!/n^n$. This limit exists; whence $\lim \root n\of{a_n}$ exists and is equal to $\lim {a_{n+1}\over a_n}$. – David Mitra Oct 07 '15 at 10:14
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I suggest you rewrite the function as $$ \lim_{ n \to \infty} e^{\frac{\log n!}{n} - \log n} $$ Then use the integral approximation for $\sum_{k=1}^{n} \log k$ to get the result $(e^{-1} )$.
Alex
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