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Question: Prove that $$\lim_{n\to\infty}(n!)^\frac1n=+\infty.$$

Solution 1: Let $\{a_n\}_{n\ge 1}$ be such that $$a_n:=(n!)^\frac{1}{n}, \forall n\in\Bbb N.$$ Now clearly $a_n>0, \forall n\in\Bbb N$. Also, $\{a_n\}_{n\ge 1}$ is a strictly increasing sequence. This implies that either $\lim_{n\to\infty}a_n=L,$ where $L\in\mathbb{R}^+$ or $\lim_{n\to\infty}a_n=+\infty$.

Let us assume that $\lim_{n\to\infty}a_n=L,$ where $L\in\mathbb{R^+}$.

Now $\forall n\in\mathbb{N},$ $$(2n)!\ge \prod_{k=n}^{2n}k\ge n^{n+1} \\\implies \sqrt[2n]{(2n)!}\ge \sqrt[2n]{n^{n+1}}=n^{\frac{n+1}{2n}}\ge n^{\frac{n}{2n}}=n^\frac{1}{2}=\sqrt{n}.$$

Therefore, we can conclude that, $\forall n\in\mathbb{N}, a_{2n}\ge \sqrt n$.

Now we know that every subsequence of a convergent sequence converges to the same limit as that of the original sequence. Thus $$\lim_{n\to\infty}a_{2n}=L\ge \lim_{n\to\infty} \sqrt{n}=+\infty,$$ which is a contradiction to the fact that $L\in\mathbb{R^+}$.

Therefore, we must have $$\lim_{n\to\infty}a_n=+\infty.$$

Solution 2: We know that for any sequence $\{a_n\}_{n\ge 1}$ of positive numbers, if $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=l, 0\le l\le \infty, \text{ then }\lim_{n\to\infty}\sqrt[n]{a_n}=l.$$ Thus defining $a_n:=n!, \forall n\in\mathbb{N},$ we have $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{(n+1)!}{n!}=\lim_{n\to\infty}(n+1)=+\infty.$$ Thus we must have $$\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\sqrt[n]{n!}=+\infty.$$ Hence, proved.

Are these solutions correct? What are the other ways to solve the problem? I was looking for a solution using the Stolz-Cesaro Lemma and also one which uses the formal definition of $$\lim_{n\to\infty}a_n=+\infty.$$ Can someone provide me with the same?

PinkyWay
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Sanket Biswas
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3 Answers3

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An alternative approach: For $1 \leq k \leq n$ we have $n!>(k)(k)..(k) =k^{n-k}$ (where there are $n-k$ factors in the product). Hence $(n!)^{1/n} >k^{1-\frac k n}$. So $\lim \inf (n!)^{1/n} \geq \lim k^{1-\frac k n}=k$. Since $k$ is arbitrary we get $\lim (n!)^{1/n} =\infty$.

Kavi Rama Murthy
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$$\log L=\lim_{n \infty} \frac{1}{n} \sum_{r=1}^{n} \log r=\lim_{n \infty} \frac{1}{n} \sum_{r=1}^{n} \log (r/n)+\log n= \int_{0}^{1} \log x dx+\infty= \infty$$ $$\implies L= \infty .$$

Z Ahmed
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$$ n!>n\times (n-1)\times \cdots \times ({n\over 2}+1)>({n\over 2})^{n\over 2} $$ hence $$ (n!)^{1\over n}>\sqrt{n\over 2} $$ Done!

Mostafa Ayaz
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