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How to prove that$(n!)^{1/n}$ tends to infinity as limit tends to infinity? I tried to do this by expanding $n!$ as $n\times (n-1)\times (n-2)\cdots 4\times3\times2\times 1$ and taking out n common from each factor so that I can have $n$ outside the radical sign, But then the last terms would be $(4/n)\times(3/n)\times(2/n)\times (1/n)$, which would tend to zero and would present indeterminate form of $0\cdot \infty$, but how should I further solve it. I would appreciate a little help.

David Holden
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user247855
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    Are you familiar with Stirling's approximation? https://en.wikipedia.org/wiki/Stirling%27s_approximation – Paul Aug 11 '15 at 15:08
  • Have you tried working with the definition? $a_n\rightarrow \infty (n\rightarrow \infty)$ if $\forall M\in \mathbb{R}, \exists N\in \mathbb{N}, \forall n\ge N\Rightarrow a_n>M$ – Anthony Aug 11 '15 at 15:14

4 Answers4

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Hint: for any $a > 1$, we have $n! > a^n$ for sufficiently large $n$.

Ben Grossmann
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Denote $(n!)^{1/n}$ by $a_n$, then $$\log a_n = \frac{1}{n}\log n! = \frac{\log 1 + \log 2 + \cdots + \log n}{n}.$$ By the celebrated Cesaro's theorem (note the result also holds if the general term tends to $\infty$), since $\log n \to \infty$ as $n \to \infty$, we have $\log a_n \to \infty$ as $n \to \infty$. Consequently, $a_n \to \infty$ as $n \to \infty$.

Zhanxiong
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Using Sterling:

$$\sqrt{2\pi}\ n^{n+1/2}e^{-n} \le n! \le e\ n^{n+1/2}e^{-n} $$

Lets apply $^{\frac{1}{n}}$

$$(\sqrt{2\pi}\ n^{n+1/2}e^{-n})^{\frac{1}{n}} \le (n!)^{\frac{1}{n}} \le (e\ n^{n+1/2}e^{-n})^{\frac{1}{n}} $$

The left side $$\sqrt[2n]{2\pi}\ n^{1+\frac{1}{2n}}e^{-1} \le (n!)^{\frac{1}{n}}$$ So:

$$\sqrt[2n]{2\pi n}\ e^{-1} n \le (n!)^{\frac{1}{n}}$$

and

$$\lim_{n\to \infty} \sqrt[2n]{2\pi n}\ e^{-1} n \to \infty$$

3d0
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$$ \lim_{n \to \infty} (n!)^{\frac1n} = \lim_{n \to \infty} \exp\left({\frac1n}\sum_{k=1}^n \log k\right) $$ for any $n \gt 1$ we have $$ n\log n -n+1 = \int_1^n \log x dx \lt \sum_{k=1}^n \log k\ \lt \int_1^n \log (x+1) dx \\= (n+1)\log(n+1) -(n+1) -2\log 2 +2 $$ i.e. $$ \log \frac{n}{e} +\frac1{n} \lt \frac1{n} \sum_{k=1}^n \log k \lt (1 +\frac1n)\left(\log n + \log(1 + \frac1n) \right) - 1 + \frac1n -\frac{2\log2}n \\ = \log\frac{n}{e}+\frac1n+\frac{\log n}n +O(\frac1n) $$ taking exponentials and dividing through: $$ e^{\frac1n} \lt \frac{(n!)^{\frac1n}}{\frac{n}{e}} \lt (Ken)^{\frac1n} $$ for a positive constant $K$

thus, by squeeze, as $n$ becomes large $$ \frac {(n!)^{\frac1n}}{ \frac{n}e} \to 1 $$

David Holden
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