Monotonicity of $\frac{n}{\sqrt[n]{(n!)}}$

Solution 1:

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).

Solution 2:

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 $e^{1-n} \leq \frac {n!}{n^n}$

Solution 3:

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 $\forall x>0\,-\mathrm{ln}2<f(x)<0$.

Solution 4:

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 $\frac{d}{dx}f^{1/x}>0$.