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I saw this question and I quickly noticed that the first inequality (after using $e^x$ on both sides) $\frac{N^N}{e^N}< N!$ can be written as $\frac{N!}{N^N}> e^{-N}$ or $\left(\frac{N!}{N^N}\right)^\frac{1}{N}> \frac{1}{e}$. This I want to prove in the end.

The famous limit $\lim\limits_{N \to \infty}\left(\frac{N!}{N^N}\right)^\frac{1}{N}= \frac{1}{e}$ will play a role in a purely algebraic proof, because if I can prove that the sequence $a_N:=\left(\frac{N!}{N^N}\right)^\frac{1}{N} $ is monotone decreasing, this will mean that $a_N > \frac{1}{e}$. But I couldn't find a proof. I tried to use the AM-GM inequality to prove that $\frac{\left(1+\frac{1}{n}\right)^{n}}{n} > \sqrt[n]{\frac{1}{n!}}$ and failed because it will result in a much more complicated problem.

The graphing calculator suggests that this sequence is indeed monotone decreasing (for all $x$). enter image description here

Although Mittens gave a good proof to the linked question, which of course counts as a proof to this question, I wonder if there is a purely algebraic proof that the sequence $a_N:=\left(\frac{N!}{N^N}\right)^\frac{1}{N} $ is monotone decreasing without using integrals.

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pie
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  • Algebraic is more aboutt identities; Calculus is probably when one needs (differential if you want to avoid integration) – Mittens Jan 28 '24 at 02:44
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    It is interesting to notice that if $A_n=\frac{n!}{n^n}$, $$\lim_n\sqrt[n]{A_n}=\lim_n\frac{A_{n+1}}{A_n}=\Big(1+\frac1n\Big)^{-n}$$ and that $(1+1/n)^{-n}\searrow e^{-1}$. Also, $A_{n+1}<A_n$ ans so, $A^{1/n}_{n+1}<A^{1/n}_n$. – Mittens Jan 28 '24 at 03:00
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    Does this help? It reduces to the problem of showing that the sequence $$\left( {1 + \frac{1}{n}} \right)^n$$ is increasing which can be done via AM-GM. – Gary Jan 28 '24 at 03:21
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    @pie: It seems that your question is answered in the positive here using induction. – Mittens Jan 28 '24 at 05:06
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    @Mittens This is what I linked as well. – Gary Jan 28 '24 at 07:01

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Yes, there's a purely algebratic proof for that.

Lemma 1. $$(n+2)\ln(n+1)-(n+1)\ln(n+2)<n\ln n-(n-1)\ln(n+1)$$ Proof. By AM-GM we have $$(n+1)^nn<\left(\frac{n(n+1)+n}{n+1}\right)^{n+1}=\left(\frac{n^2+2n}{n+1}\right)^{n+1}$$ $$\ln n+n\ln(n+1)<(n+1)\ln(n^2+2n)-(n+1)\ln(n+1)$$ $$\ln n+n\ln(n+1)<(n+1)\ln(n+2)+(n+1)\ln n-(n+1)\ln(n+1)$$ $$(2n+1)\ln(n+1)<(n+1)\ln(n+2)+n\ln n$$ QED.

Lemma 2. $$n(n+1)\ln n-n^2\ln(n+1)<\ln(n!)$$ Proof. By mathematical induction:

When $n=1$, $-\ln2<0$ is true.

Assume it's true for $n$, then $$ \begin{aligned} &(n+1)(n+2)\ln(n+1)-(n+1)^2\ln(n+2)\\ =&(n+1)((n+2)\ln(n+1)-(n+1)\ln(n+2))\\ <&(n+1)(n\ln n-(n-1)\ln(n+1))\\ =&n(n+1)\ln n-n^2\ln(n+1)+\ln(n+1)\\ <&\ln(n!)+\ln(n+1)\\ =&\ln((n+1)!) \end{aligned} $$ so it's true for $n+1$.

QED. $$n(n+1)\ln n-n^2\ln(n+1)<\ln(n!)$$ $$n\ln(n!)-n^2\ln(n+1)<(n+1)\ln(n!)-n(n+1)\ln n$$ $$\frac{1}{n+1}\ln(n!)-\frac{n}{n+1}\ln(n+1)<\frac{1}{n}\ln(n!)-\ln n$$ $$\frac{1}{n+1}(\ln(n!)-n\ln(n+1))<\frac{1}{n}(\ln(n!)-n\ln n)$$ $$\left(\dfrac{n!}{(n+1)^n}\right)^{\frac{1}{n+1}}<\left(\dfrac{n!}{n^n}\right)^{\frac{1}{n}}$$ $$\left(\dfrac{(n+1)!}{(n+1)^{n+1}}\right)^{\frac{1}{n+1}}<\left(\dfrac{n!}{n^n}\right)^{\frac{1}{n}}$$ Which is $a_{n+1}<a_n$.

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