Infinite series of nth root of n factorial

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)

Solution 1:

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?

Solution 2:

An easy way to approach it is Stirling's approximation: $n! \approx (\frac ne)^n\sqrt{2 \pi n}$ so $n!^{\frac 1n} \approx \frac ne \to \infty$

Solution 3:

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 $\infty$.