Monotonicity of $\frac{n}{\sqrt[n]{(n!)}}$ It is known that when $n\rightarrow\infty$ the sequence $$\frac{n}{\sqrt[n]{(n!)}}$$ has limit $e$ but I don't know how to prove it's monotonicity. After a short calculus using WolframAlpha I found that this sequence is actually increasing. I tried to compare $2$ consecutive members but I couldn't manage to show something. It is obvious that $$\frac{n^n}{n!}$$ is increasing but that don't help us much (I think)
 A: Taking $\log$ gives $$\sum_{i=1}^n\frac{-\log \frac i n}{n}$$
This is a Riemann sum of the integral $\int_{0}^1(-\log x)\,dx$.
Since the function $-\log x$ is decreasing and convex, its Riemann sums have to increase monotonically as the partition gets finer, if the partitions are regular. (Proof here).

A: We need to show that
$$a_n=\frac{n}{\sqrt[n]{n!}}\quad a_{n+1}\geq a_n\iff$$ 
$$\frac{n+1}{\sqrt[n+1]{(n+1)!}}\geq \frac{n}{\sqrt[n]{n!}}
\iff 
\frac{n+1}{ (n+1)!^ \left(\frac{1}{n+1}\right) }n!^\left( \frac{1}{n} \right)\geq n \iff$$
$$ (n+1)^{\left(1-\frac{1}{n+1}\right)}n!^{\left(\frac{1}{n} -\frac{1}{n+1}\right)}\geq n 
\iff 
(n+1)^{\left(\frac{n}{n+1}\right)}n!^{\left(\frac{1}{n(n+1)}\right)}\geq n $$
$$
(n+1)^{n^2}n!\geq n^{\left(n(n+1)\right)}=n^{n^2}n^n\iff \frac{n^n}{n!}\leq \left(\frac{n+1}{n}\right)^{n^2}=\left(1+\frac1n\right)^{n^2}\iff $$
$$\left(1+\frac1n\right)^{n^2} \geq\frac{n^n}{n!}$$
which is true since
$$\left(1+\frac1n\right)^{n^2}\geq \left(1+\frac1n\right)^{n^2-1}=\left[\left(1+\frac1n\right)^{n+1}\right]^{n-1}\geq e^{n-1}\geq\frac{n^n}{n!}$$
indeed
$$e^{n-1}\geq\frac{n^n}{n!}\iff b_n=\frac {e^nn!}{n^n}\geq e$$
which is true $\forall n$ since
$$n=1 \implies b_1=\frac{e^11!}{1^1}\geq e$$
and
$$\frac{b_{n+1}}{b_n}=\frac {e^{n+1}(n+1)!}{(n+1)^{n+1}}\frac {n^n}{e^nn!}=\frac{e}{\left(1+\frac1n\right)^n}>1$$
For the last inequality see also the related OP
Show that $e^{1-n} \leq \frac {n!}{n^n}$
A: We have that $\frac{n}{(n!)^{\frac{1}{n}}}\leq\frac{n+1}{((n+1)!)^{\frac{1}{n+1}}}\Leftrightarrow\frac{n^{n(n+1)}}{(n!)^{n+1}}\leq\frac{(n+1)^{n(n+1)}}{((n+1)!)^{n}}\Leftrightarrow$ simplifying, $\frac{n^{n^2+n}}{n!}\leq(n+1)^{n^2}\Leftrightarrow\frac{n^{n^2+n}}{(n!)(n+1)^{n^2}}\leq1$;let us proceed by induction. For n=1 $\frac{1}{2}\leq1\Rightarrow$ it is true. If it is true for n, for n+1 we have $\frac{(n+1)^{(n+1)^2+n+1}}{((n+1)!)(n+1+1)^{(n+1)^2}}=\frac{(n+1)^{n^2+3n+1}}{(n!)(n+2)^{(n+1)^2}}=\frac{n^{n^2+n}}{(n!)(n+1)^{n^2}}\cdot\frac{(n+1)^{(n+1)^2+n^2+n}}{n^{n^2+n}(n+2)^{(n+1)^2}}\leq\frac{(n+1)^{(n+1)^2+n^2+n}}{n^{n^2+n}(n+2)^{(n+1)^2}}$, by the induction hypothesis. So, if we show that $\frac{(n+1)^{(n+1)^2+n^2+n}}{n^{n^2+n}(n+2)^{(n+1)^2}}\leq1$, then we have proved the induction step, and thus proved the thesis. Now, this last statement holds $\Leftrightarrow (n+1)^{(n+1)^2+n^2+n}=(n+1)^{(n+1)(2n+1)}\leq n^{n^2+n}(n+2)^{(n+1)^2}=n^{n(n+1)}(n+2)^{(n+1)^2}\Leftrightarrow(n+1)^{2n+1}\leq n^n(n+2)^{n+1}$ (because $n,n+1,n+2\geq1$), $\Leftrightarrow f(n)\leq0$, with
$f:(0,+\infty)\rightarrow\mathbb{R},x\mapsto (2x+1)\mathrm{ln}(x+1)-x\mathrm{ln}(x)-(x+1)\mathrm{ln}(x+2)$.
We have that $f(x)\to-\mathrm{ln}2<0$ as $x\to0^+$ (because $x\mathrm{ln}x\to0$) and $f(x)=x\mathrm{ln}\frac{x+1}{x}+(x+1)\mathrm{ln}\frac{x+1}{x+2}=x\mathrm{ln}(\frac{x+1}{x}\cdot\frac{x+1}{x+2})+\mathrm{ln}\frac{x+1}{x+2}$, where the second term tends to 0, and if $y=\frac{x+1}{x}\cdot\frac{x+1}{x+2}-1=\frac{1}{x(x+2)}>0\Leftrightarrow yx^2+2yx-1=0\Leftrightarrow x=\frac{\sqrt{y^2+y}-y}{y}$ (we have taken the + sign because we want x>0), $y\to0$ as $x\to+\infty$, and $\lim_{x\to+\infty}x\mathrm{ln}(\frac{x+1}{x}\cdot\frac{x+1}{x+2})=(\frac{\sqrt{y^2+y}-y}{y})\mathrm{ln}(1+y)=\frac{1}{\sqrt{y^2+y}+y}\mathrm{ln}(1+y)=\frac{1}{\sqrt{1+\frac{1}{y}}+1}\frac{\mathrm{ln}(1+y)}{y}\to0\cdot1=0$ as $x\to+\infty\Leftrightarrow y\to0^+$.
Next, we compute f' and f''.
$f'(x)=2\mathrm{ln}(x+1)+\frac{2x+1}{x+1}-\mathrm{ln}x-1-\mathrm{ln}(x+2)-\frac{x+1}{x+2}$, and $f''(x)=\frac{-4-3x}{x(x+1)^2(x+2)^2}<0$ for x>0, and so then f' is strictly decreasing for x>0. We have that $f'(x)\to+\infty$ as $x\to0^+$,$f'(x)\to0$ as $x\to+\infty$, so $f'(x)>0$ for x>0, and so $\forall x>0\,-\mathrm{ln}2<f(x)<0$.
A: *

*First method


Consider $n!=\Gamma(n-1)$, where $\Gamma$ is the analytical extension of the factorial to complex numbers . Your expression makes now sense even if $n$ is real, and thus you can derivate (our expression, restricted to $\mathbb{R}^+$ is $C^{\infty}$). By your own consideration, we have that:
$f(x)=\frac{x^x}{\Gamma(x-1)}$ is increasing. Thus our problem reduces to showing that $\frac{d}{dx}f^{1/x}>0$.


*

*Second method


To proove that your expression converges to $e$, you can use the Stirling's approximation:
$\begin{equation}
n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n
\end{equation}$
(where $~$ means that the their ratio tend to $1$ as $n \to \infty$)
Thus, your expression becomes:
$\begin{align}
\lim_{n \to \infty}\frac{n}{(2πn)^{\frac{2}{n}}\frac{n}{e}}=\\
\lim_{n \to \infty} \frac{e}{(2πn)^{\frac{2}{n}}}=\\
e
\end{align}$
And the second equation shows that, for $n$ l'arte enough, the expression is increasing
A: $a_n
=\frac{n}{\sqrt[n]{(n!)}}
$.
$b_n
=\ln a_n
=\ln n-\ln(n!)/n
$.
Plunging ahead
naively,
and using
$\ln n!
\gt n\ln n-n +\frac12 \ln n+c
$
where
$c 
= \frac12 \ln(2\pi)
\approx 0.919$
(from
https://en.wikipedia.org/wiki/Stirling%27s_approximation)
$\begin{array}\\
b_{n+1}-b_n
&=\ln (n+1)-\ln((n+1)!)/(n+1)-\ln n+\ln(n!)/n\\
&=\ln (1+1/n)-(\ln(n!)+\ln(n+1))/(n+1)+\ln(n!)/n\\
&=\ln (1+1/n)-\ln(n+1)/(n+1)+\ln(n!)(1/n-1/(n+1))\\
&=\ln (1+1/n)-\ln(n+1)/(n+1)+\ln(n!)(1/(n(n+1))\\
&>\ln (1+1/n)-\ln(n+1)/(n+1)+(n\ln n-n+c+\frac12 \ln n)/(n(n+1))\\
&=\ln (1+1/n)+(\ln n-\ln(n+1)/(n+1)-1/(n+1)+(c+\frac12 \ln n)/(n(n+1))\\
&>\ln (1+1/n)-\ln(1+1/n)/(n+1)-1/(n+1)+(c+\frac12 \ln n)/(n(n+1))\\
&=(1-1/(n+1))\ln (1+1/n)-1/(n+1)+(c+\frac12 \ln n)/(n(n+1))\\
&>(n/(n+1))(1/n-1/(2n^2))-1/(n+1)+(c+\frac12 \ln n)/(n(n+1))\\
&>(n/(n+1))((2n-1)/(2n^2))-1/(n+1)+(c+\frac12 \ln n)/(n(n+1))\\
&=(2n-1)/(2n(n+1))-1/(n+1)+(c+\frac12 \ln n)/(n(n+1))\\
&=(-1)/(2n(n+1))+(c+\frac12 \ln n)/(n(n+1))\\
&=(c-\frac12+\frac12 \ln n)/(n(n+1))\\
&>\ln n/(2n(n+1))\\
&> 0\\
\end{array}
$
Note:
According to Wolfy,
this is correct.
A: The ratio of the consecutive terms of the sequence $a_n=\frac{n}{\sqrt[n]{n!}}$ is greater than $1$.
Therefore, $\frac{n}{\sqrt[n]{n!}}$ is increasing.
$$
\begin{align}
\left(\frac{a_{n+1}}{a_n}\right)^{n(n+1)}
&=\left[\frac{\frac{\color{#C00}{n+1}}{\color{#090}{\sqrt[n+1]{(n+1)!}}}}{\frac{\color{#C00}{n}}{\color{#090}{\sqrt[n]{n!}}}}\right]^{n(n+1)}\tag1\\
&=\left(\color{#C00}{\frac{n+1}n}\right)^{n(n+1)}\color{#090}{\frac{n!^{n+1}}{(n+1)!^n}}\tag2\\[6pt]
&=\left(\frac{n+1}n\right)^{n(n+1)}\frac{n!}{(n+1)^n}\tag3\\[6pt]
&=\left(\frac{n+1}n\right)^{n^2}\frac{n!}{n^n}\tag4\\[15pt]
&\ge2\tag5
\end{align}
$$
Explanation:
$(2)$: distribute the exponent $n(n+1)$ over the ratios
$(3)$: cancel $n!^n$ in the numerator and denominator of the right term
$(4)$: move $\left(\frac{n+1}n\right)^n$ from the left term to the right term
$(5)$: inequality $(10)$ shows that $b_n=\left(\frac{n+1}n\right)^{n^2}\frac{n!}{n^n}$ is increasing
$\phantom{\text{(5):}}$ since $b_1=2$, $b_n\ge2$

Bernoulli's Inequality shows that $b_n=\left(\frac{n+1}n\right)^{n^2}\frac{n!}{n^n}$ is increasing:
$$
\begin{align}
\frac{b_{n+1}}{b_n}
&=\frac{\color{#C00}{\left(\frac{n+2}{n+1}\right)^{(n+1)^2}}\color{#090}{\frac{(n+1)!}{(n+1)^{n+1}}}}{\color{#C00}{\left(\frac{n+1}n\right)^{n^2}}\color{#090}{\frac{n!}{n^n}}}\tag6\\
&=\color{#C00}{\left(\frac{n(n+2)}{(n+1)^2}\right)^{(n+1)^2}\left(\frac{n+1}n\right)^{2n+1}}\color{#090}{\left(\frac{n}{n+1}\right)^n}\tag7\\
&=\left(1-\frac1{(n+1)^2}\right)^{(n+1)^2}\left(\frac{n+1}n\right)^{n+1}\tag8\\[6pt]
&\ge\left(1-\frac1{n+1}\right)^{n+1}\left(\frac{n+1}n\right)^{n+1}\tag9\\[12pt]
&=1\tag{10}
\end{align}
$$
Explanation:
$\phantom{1}(7)$: multiply numerator and denominator by $\left(\frac{n+1}n\right)^{2n+1}$ and combine terms
$\phantom{1}(8)$: $\frac{n(n+2)}{(n+1)^2}=1-\frac1{(n+1)^2}$
$\phantom{1}(9)$: Bernoulli's Inequality
$(10)$: $1-\frac1{n+1}=\frac{n}{n+1}$
