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Let $n$ be a positive integer such that $n>1$. Then, how do I prove that $n!\leq (n/\sqrt{2})^n$?

Let $K$ be a number field and $s$ be the number of non-real embeddings $K\rightarrow \mathbb{C}$.

Then, there exists a constant $M_K$ such that for any nonzero ideal $I$ of $O_K$, there exists a nonzero element $a\in I$ such that $|N_{K/\mathbb{Q}}(a)|\leq M_K \sqrt{|\Delta_K|}N(I)$.

$M_K$ can be taken to be $(\frac{4}{\pi})^s \frac{n!}{n^n}$, and this number is called the Minkowski bound.

However, Neukirsch, in his text, proves the above theorem with $M_K=(\frac{2}{\pi})^s$.

I'm not sure which one is smaller. Note that Neukirsch's $M_K$ is greater than the Minkowski's $M_K$ if we can prove that $n!\leq (n/\sqrt{2})^n$ for any positive integer $n>1$, but I am not sure how to prove this.

Rubertos
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    Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Fly by Night Aug 31 '17 at 00:36
  • Logarithm may do? – BAI Aug 31 '17 at 00:36
  • Well, I edited my question. However, I am not sure how this provides more information on this problem. – Rubertos Aug 31 '17 at 00:46

4 Answers4

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The case $n=2$ can be checked separately. For $n\ge3,$ by the AM-GM inequality we have $$(n!)^{\frac1n}\le\frac{1+2+\cdots+n}n=\frac{n+1}2\lt\frac n{\sqrt2},$$ so $$n!\lt\left(\frac n{\sqrt2}\right)^n.$$

bof
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Wikipedia gives $ n! < (n/2)^n$ for $n\ge 6$, which implies $n!\leq (n/\sqrt{2})^n$.

The cases $2 \le n \le 5$ can be done manually.

Using the better estimate $$ n! \leq e\left(\frac{n+1}e\right)^{n+1} $$ it suffices to do the cases $2 \le n \le 3$ manually because $$ e\left(\frac{n+1}e\right)^{n+1} \le \left(\frac{n}{\sqrt{2}}\right)^n $$ for $n \ge 4$.

lhf
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It is very easy to see by looking at $(n!)^2$ and pairing small terms with large terms, that $$n! \le (\tfrac{n+1}{2})^n$$ for all $n\ge 1$. We then easily verify that $(n+1)/2 < n/\sqrt2$ for all $n \ge 3$, leaving only the case $n=2$ to check separately.

Erick Wong
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A most elementary approach: Note that if $n \geq 2$ is even then \begin{align*} n! & = \bigg(1 \cdots \frac{n}{2}\bigg)\bigg(\big(\frac{n}{2}+1\big) \cdots n\bigg) \\ & \leq \bigg( \frac{n}{2} \cdots \frac{n}{2} \bigg) \bigg( n \cdots n \bigg) \\ &= n^n / 2^{n/2} \end{align*}

If $n>2$ is odd then $n+1<\sqrt 2 n$, so similarly we get \begin{align*} n! &= \bigg(1 \cdots \frac{n-1}{2}\bigg)\cdot \frac{n+1}{2}\cdot \bigg(\big(\frac{n+1}{2}+1\big) \cdots n\bigg) \\ & \leq \bigg( \frac{n}{2} \cdots \frac{n}{2} \bigg)\cdot \frac{n}{\sqrt 2} \cdot \bigg( n \cdots n \bigg) \\ &= n^n / 2^{n/2} \end{align*}

shalop
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