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)

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