TITLE: Infinite series of nth root of n factorial
QUESTION [11 upvotes]: Why is this not correct:
$$
\begin{align}
\lim_{n\to \infty}\sqrt[n]{n!} &=
\lim_{n\to \infty}\sqrt[n]{n(n-1)(n-2)(n-3)\cdots(1)} \\
&=\lim_{n\to \infty}\sqrt[n]{n} \cdot \lim_{n\to \infty}\sqrt[n]{n-1} \cdot \lim_{n\to \infty}\sqrt[n]{n-2}\cdots \lim_{n\to \infty}\sqrt[n]{1} \\
&=1 \cdot 1 \cdot 1 \cdot 1 \cdots 1 \\
&=1
\end{align}
$$
Therefore, $\lim_{n\to \infty} \sqrt[n]{n!}=1$.
It is clear that $\lim_{n\to \infty} \sqrt[n]{n}= 1$ as and that $n! = n(n-1)!$
Yet wolframalpha gives me infinity as the limit and not $1$!
If you have Rudin's Principles of Mathematical Analysis refer to Theorem $3.3$ c) and Theorem $3.20$ c)

REPLY [2 votes]: Let $(a_n)$ be a pozitive sequence. If $\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}$  and $\lim_{n \to \infty}\sqrt[n]{a_n}$ both  exists then:  $\lim_{n \to \infty} \frac {a_{n+1}}{a_n}=\lim_{n \to \infty}\sqrt[n]{a_n}$ Apply this for $n!$ 
Actually for a pozitive sequence $(a_n) $ we have: $\liminf\frac{a_{n+1}}{a_n}\leq\liminf\sqrt[n]{a_n}\leq\limsup\sqrt[n]{a_n}\leq\limsup\frac{a_{n+1}}{a_n}$
Furthermore limit of a sequence $(a_n)$ exists iff $\limsup a_n=\liminf a_n$