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Show that $\dfrac{(m!)^{1/m}}{m/e}$ is a decreasing function of $m$.

Here is my proof.

I would like to see others, preferably simpler.

I have shown in Proof explanation $\lim\limits_{n\to\infty}\frac{n}{\sqrt[n]{n!}}=e$ that, if $r_m =\dfrac{(m!)^{1/m}}{m/e} $, then $r_m > 1+1/m $.

This was based on $(1+1/m)^m < e \lt (1+1/m)^{m+1} $, which was also shown there.

Since $r_m \to 1$, this leads to the conjecture that $r_m$ is a decreasing function of $m$.

Let $s_m = \dfrac{r_{m+1}}{r_m}$.

Then, since $r_m > 1+1/m$ implies that $m! > (1+1/m)^mm^m/e^m$,

$\begin{array}\\ s_m &= \dfrac{r_{m+1}}{r_m}\\ &= \dfrac{\dfrac{((m+1)!)^{1/(m+1)}}{(m+1)/e}}{\dfrac{(m!)^{1/m}}{m/e}}\\ &= \dfrac{\dfrac{((m+1)!)^{1/(m+1)}}{(m+1)}}{\dfrac{(m!)^{1/m}}{m}}\\ &= \dfrac{m((m+1)!)^{1/(m+1)}}{(m+1)(m!)^{1/m}}\\ \text{so}\\ s_m^{m(m+1)} &= \dfrac{m^{m(m+1)}((m+1)!)^{m}}{(m+1)^{m(m+1)}(m!)^{m+1}}\\ &= \dfrac{m^{m(m+1)}m!^m(m+1)^m}{(m+1)^{m(m+1)}m!^{m}m!}\\ &= \dfrac{m^{m(m+1)}}{(m+1)^{m^2}m!}\\ &= \dfrac{m^{m}}{(1+1/m)^{m^2}m!}\\ &< \dfrac{m^{m}}{(1+1/m)^{m^2}(1+1/m)^mm^m/e^m}\\ &= \dfrac{e^m}{(1+1/m)^{m^2+m}}\\ \text{so}\\ s_m^{m+1} &< \dfrac{e}{(1+1/m)^{m+1}}\\ &< 1\\ \end{array} $

I find it pleasing that the proof ends up depending on the upper bound for $e$.

user26857
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marty cohen
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1 Answers1

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I guess the following is "quicker", but relies on some heavy machinery. We have \begin{align*} \frac{d}{dm} \log r_m &= -\frac{1}{m^2} \log m! + \frac{1}{m} \frac{\Gamma'(m+1)}{m!} - \frac{1}{m} \\ &\le -\frac{\log\sqrt{2\pi}-m+(m+\frac{1}{2})\log m}{m^2} + \frac{H_m - \gamma - 1}{m} \end{align*} where we used the fact that $\sqrt{2\pi}m^{m+\frac{1}{2}}e^{-m} \le m!$ and $\Gamma'(m+1) = m!(-\gamma + H_m)$, where $H_m$ is the $m$th harmonic number. So \begin{align*} \frac{d}{dm} \log r_m &\le \frac{1}{m}\left(H_m - \left(1 + \frac{1}{2m}\right)\log m - \gamma - \frac{1}{m}\log\sqrt{2\pi}\right) \end{align*} But since \begin{align*} H_m &= \log m + \gamma + \frac{1}{2m} - \frac{1}{12m^2} + \cdots \\ &< \log m + \gamma + \frac{1}{2m} \\ &< \left(1 + \frac{1}{2m}\right)\log m + \gamma + \frac{1}{m}\log\sqrt{2\pi} \end{align*} we are done.

Tom Chen
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