Is $(n+1)! = \mathrm{o}(n^n)$?

Question

Continuing http://math.stackexchange.com/a/2431263, it seems to me that you can prove with the same method that $(n+1)! = o(n^n)$.

Namely:

Let $a_n=\frac{(n+1)!}{n^n}$. Then:

$$\frac{a_{n+1}}{a_n}=\frac{\frac{(n+2)!}{(n+1)^{n+1}}}{\frac{(n+1)!}{n^n}}=\frac{n^n(n+2)!}{(n+1)!(n+1)^{n+1}}=\frac{n^n}{(n+1)^n}\frac{n+2}{n+1}=\left(\frac{n}{n+1}\right)^n \left(1+\frac{1}{n+1}\right) \to \frac{1}{e}\cdot 1 <1$$

Thus from ratio test for sequences we have that $a_n \to 0$.

Can anyone confirm or reject this?

Answer 1

Compare

$$LHS=\color{green}2\cdot3\cdot4\cdots n\cdot\color{green}{(n+1)}$$ and $$RHS=\color{green}n\cdot n\cdot n\cdot n\cdots \color{green}n$$ ($n$ factors each).

Now it is clear that $LHS=o( RHS)$ just because $\color{green}2\cdot\color{green}{(n+1)}=o(\color{green}{n\cdot n})$ and all intermediate factors are smaller or equal.

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More generally, for any constant $k$, $(n+k)!=o(n^n)$.

Answer 2

Your answer is close, but it seems more satisfying to use $\frac{n+2}{n}\frac1{\left(1+\frac1n\right)^{n+1}}$ as the ratio because it gives telescoping inequalities.

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Let $$ a_n=\frac{(n+1)!}{n^n}\tag1 $$ then, because $\frac{n+2}{n}\le\left(\frac{n+1}{n}\right)^2$ (expand the square) and $\left(1+\frac1n\right)^{n+1}\ge e$ (see this answer, which says that $\left(1+\frac1n\right)^{n+1}$ decreases to $e$), $$ \begin{align} \frac{a_{n+1}}{a_n} &=(n+2)\frac{n^n}{(n+1)^{n+1}}\\ &=\frac{n+2}{n}\frac1{\left(1+\frac1n\right)^{n+1}}\\ &\le\left(\frac{n+1}{n}\right)^2\frac1e\tag2 \end{align} $$ Therefore, telescopically, we have $$ a_n\le a_1\frac{n^2}{e^{n-1}}\tag3 $$ and since $$ \lim_{n\to\infty}\frac{n^2}{e^{n-1}}=0\tag4 $$ we get that $(n+1)!=o\!\left(n^n\right)$