\[\sqrt[n]{n!} \geq \sqrt{n}\]

Prove the above for all positive integers \(n\).

- Try to give a non-inductive, non-calculus proof

\[\sqrt[n]{n!} \geq \sqrt{n}\]

Prove the above for all positive integers \(n\).

- Try to give a non-inductive, non-calculus proof

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TopNewestSo I will give an inductive, calculus-requiring proof.

Lemma 1: \(\left( 1 + \frac{1}{n} \right)^n \le e\) for all positive integer \(n\).One way to prove it is to observe that \(1 + \frac{1}{n} \le 1 + \frac{2}{2n} + \frac{1}{4n^2} = \left( 1 + \frac{1}{2n} \right)^2\), thus by induction,

\[\displaystyle\begin{align*} \left(1 + \frac{1}{n} \right)^n &\le \left(1 + \frac{1}{2n} \right)^{2n} \\ &\le \left(1 + \frac{1}{4n} \right)^{4n} \\ &\le \ldots \\ &\le \lim_{k \to \infty} \left(1 + \frac{1}{2^k n} \right)^{2^k n} \\ &= \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n \\ &= e \end{align*}\]

Lemma 2: \(e < 3\)\[\displaystyle\begin{align*} e &= \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \ldots \\ &< 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \ldots \\ &= 1 + 2 \\ &= 3 \end{align*}\]

Now, the claim \(\sqrt[n]{n!} \ge \sqrt{n}\) is equivalent to \((n!)^2 \ge n^n\). We will prove this by induction. The base cases are \(n = 1, 2, 3\) in which it is true. Now suppose that it's true for some \(n \ge 3\), and we'll prove the claim for \(n \gets n+1\).

\[\displaystyle\begin{align*} ((n+1)!)^2 &= (n!)^2 \cdot (n+1)^2 \\ &\ge n^n \cdot (n+1)^2 \\ &\ge n^n \cdot (n+1) \cdot e & \text{by Lemma 2, because $n \ge 3 > e$} \\ &\ge n^n \cdot (n+1) \cdot \left( 1 + \frac{1}{n} \right)^n & \text{by Lemma 1} \\ &= \left(n + \frac{n}{n} \right)^n \cdot (n+1) \\ &= (n+1)^n \cdot (n+1) \\ &= (n+1)^{n+1} \end{align*}\] – Ivan Koswara · 1 year ago

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We have to prove \((n!)^2 \ge n^n\). Instead of proving it using Sterling's approximation or induction here is a solution on AM-GM. I found a nice form of \((n!)^2\) recently. \((n!)^2=[1 \cdot n][2 \cdot (n-1)]\ldots[n \cdot1]\)(there are \(n\) terms). Now, for some variable \(m\), such that \(1 \le m \le n\) we have \(m(n-m+1) \ge n\). The result follows. – Svatejas Shivakumar · 1 year ago

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– Nihar Mahajan · 1 year ago

Wise solution!Log in to reply

\[1\leq m\leq n\] is \[ m(n-m+1)\geq n\] ? – Upamanyu Mukharji · 1 year ago

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– Svatejas Shivakumar · 1 year ago

Take \(n=m+k\) where \(k \ge 0\). The inequality simplifies to \(m(k+1) \ge m+k\) or \(mk+m \ge m+k\) or \(m \ge 1\) which is true.Log in to reply

– Upamanyu Mukharji · 1 year ago

OK.. Thanks!Log in to reply

– Anirudh Chandramouli · 1 year ago

Nice solutionLog in to reply

\(\sqrt[n]{n!} \geq \sqrt{n}\)

\({n!} \geq {\sqrt{n}}^n\)

\((n) \cdot (n-1) \cdot (n-2) \cdot \dots \cdot (1) \geq {\sqrt{n}}^n\)

\((n)(1) \cdot (n-1)(1+1) \cdot (n-2)(1+2) \cdot \dots \cdot (\frac{n+1}{2})(\frac{n-1}{2}) \geq {\sqrt{n}}^n\) or \((n)(1) \cdot (n-1)(1+1) \cdot (n-2)(1+2) \cdot \dots \cdot (\frac{n}{2}) \geq {\sqrt{n}}^n\)

We note that (n-k)(1+k) > n for all -1<k<n+1. So the result follows. By the way, i was sniped by Svatejas Shivakumar. – Aloysius Ng · 1 year ago

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– Nihar Mahajan · 1 year ago

Ninja'ed haha tfsLog in to reply

Well what about n-1? – Oscar Oquelí · 1 year ago

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