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)
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$\begingroup$Consider: $\lim_{n \to \infty} 1 = \lim_{n\to \infty} (1/n + \cdots + 1/n) = \sum \lim_{n \to \infty} 1/n = \sum 0 = 0$, and compare that with what you did. Do you understand why your second "equality" isn't correct?
$\endgroup$ $\begingroup$An easy way to approach it is Stirling's approximation: $n! \approx (\frac ne)^n\sqrt{2 \pi n}$ so $n!^{\frac 1n} \approx \frac ne \to \infty$
$\endgroup$ 1 $\begingroup$I could not understand the accepted answer. But I tried to solve it by constructing an inequality, whose one side is easy to compute and that computation is enough to conclude on the other side.
For any odd natural number $n$, \begin{align*} n!\,n!&=(1\cdot 2 \cdots n)\times (1\cdot 2 \cdots n)\\ &=(1\cdot n)^2\times \{2\cdot(n-1)\}^2 \times \cdots \times \{(\frac{n+1}{2})\cdot (\frac{n+1}{2})\}^2\\ &> \Big(\frac{n}{2}\Big)^2 \times \Big(\frac{n}{2}\Big)^2 \times \cdots \Big(\frac{n}{2}\Big)^2 \,\,\Big[\Big(\frac{n+1}{2}\Big) \,\,\text{times}\Big]\\ &=\Big(\frac{n}{2}\Big)^{2\big(\frac{n+1}{2}\big)}=\Big(\frac{n}{2}\Big)^{n+1} \end{align*}Then $(n!)^{\frac{2}{n}}>\big(\frac{n}{2}\big)^{\frac{n+1}{n}} \implies (n!)^{\frac{1}{n}}>\big(\frac{n}{2}\big)^{\frac{n+1}{2n}}=\big(\frac{n}{2}\big)^{\frac{1}{2}+\frac{1}{2n}}$
Similarly, for any even natural number $n$,$$n!\,n!>\Big(\frac{n}{2}\Big)^n \,\,\Big[\text{ There will be } \frac{n}{2} \text{ terms}\Big]$$Then $(n!)^{\frac{2}{n}}>\big(\frac{n}{2}\big) \implies (n!)^{\frac{1}{n}}>\big(\frac{n}{2}\big)^{\frac{1}{2}}$
Using the fact that $\lim_{n \to \infty} n^{\frac{1}{n}}=1$, one can obtain $\lim_{n \to \infty} \big(\frac{n}{2}\big)^{\frac{1}{2n}}=1$
Thus in all cases we have $\lim_{n \to \infty}(n!)^{\frac{1}{n}}\geq\lim_{n \to \infty}\big(\frac{n}{2}\big)^{\frac{1}{2}}$
The limit in the R.H.S. diverges. Hence the desired limit (in the L.H.S.) also goes to $\infty$.
$\endgroup$ 3 $\begingroup$Your solution: The number of terms is n and n goes to infinity and the multiplication of one infinitely many times is not one.
$\endgroup$ 6 $\begingroup$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 $\limsup a_n=\liminf a_n$
$\endgroup$ 3 $\begingroup$I am sorry for reopening this thread. The sole purpose is to point out what I think is a flawed argument in one of the responses. Since I do not have enough privileges to comment on the post directly, I hope someone here can help me attract the authors attention so that it can be corrected.
In his explanation about how to solve the limit, Sameh Shenawy says,
(3) commute ln and lim and simplify the function to be sum of n terms (4) Apply L'Hopital rule term by term
When applying L'Hopital's rule to the first term we get, $$ \lim_{n\to\infty} \frac{ln(n)}{n} = \lim_{n\to\infty} \frac{1/n}{1} $$
which tends to $0$ and not $1$. Similarly, all the limits will tend to $0$ leading to $ln(L)$ to also tend to $0$.
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