Proving $n!n^s=o(n^n)$

Question

I want to prove that for any $s\geq0$, and $n\to\infty$, $n!n^s=o(n^n)$ or $$\lim_\limits{n\to\infty}\frac{n!n^s}{n^n}=0.$$ As a hint, the inequality $$\sum\limits_{k=1}^{n}\log k\leq n\log\frac{n+1}2$$ is given.

I thought that instead of proving $\lim_\limits{n\to\infty}n!n^{s-n}=0$, one could also show $\lim_\limits{n\to\infty}\log\left(n!n^{s-n}\right)=-\infty$. This would be equivalent to $\lim_\limits{n\to\infty}\log\left(n!\right)+\log\left(n^{s-n}\right)=-\infty$. Using the given inequality, it would be sufficient to prove $$\lim_\limits{n\to\infty}n\log\frac{n+1}2+\log\left(n^{s-n}\right)=-\infty\Leftrightarrow\lim_\limits{n\to\infty}n\log\frac{n+1}{2n}+s\log n=-\infty\\\Leftrightarrow\lim_\limits{n\to\infty}n\log\frac12+s\log n=-\infty.$$ What should I do now?

Answer 1

The series $$\sum_\limits{n=1}^\infty\frac{n!n^s}{n^n}$$ is convergent by the ratio test (easy exercise). Therefore $$\lim_\limits{n\to\infty}\frac{n!n^s}{n^n}=0.$$

Answer 2

It is enough to prove that $\frac{n!}{n^n}$ behaves more or less like $e^{-n}$. $\log(x)$ is a concave function on $\mathbb{R}^+$, hence by the Hermite-Hadamard inequality

$$\begin{eqnarray*} \frac{1}{2}\log(1)+\log(2)+\ldots+\log(n)+\frac{1}{2}\log(n+1) &\leq& \int_{1}^{n+1}\log(x)\,dx\\ &=& (n+1)\log(n+1)-n\end{eqnarray*}$$ that translates into $$ \log(n!)\leq \left(n+\frac{1}{2}\right)\log(n+1)-n $$ such that, by exponentiating both sides, $$ \frac{n!}{(n+1)^n}\leq \frac{\sqrt{n+1}}{e^n}$$ that is enough to prove your claim.

Answer 3

Write \begin{align} n\log\left(\frac{n+1}{2n}\right) + s \log n = n \left(\log\left(\frac{n+1}{2n}\right) + s\cdot\frac{\log n}{n}\right) \end{align} Now since $$ \log\left(\frac{n+1}{2n}\right) + s\cdot\frac{\log n}{n} \to \log\left(\frac{1}{2}\right) < 0 $$ it follows that $$ n \left(\log\left(\frac{n+1}{2n}\right) + s\cdot\frac{\log n}{n}\right) \to -\infty $$

Answer 4

Using the hint given,

\begin{align} \log \left( \frac{n! n^s}{n^n} \right) &= \sum_{k=1}^n \log(k) + \log\left(\frac{n^s}{n^n}\right)\\ &\leq n\log\left(\frac{n+1}{2}\right) + \log\left(\frac{n^s}{n^n}\right)\\ &= \log\left( \frac{(n+1)^n}{2^n} \cdot \frac{n^s}{n^n}\right)\\ &= \log\left( \left(1+\frac{1}{n}\right)^n \cdot \frac{n^s}{2^n}\right) \end{align} By exponentiation

\begin{align} 0 \leq \lim_{n\to\infty} \frac{n! n^s}{n^n} \leq \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n \cdot \frac{n^s}{2^n} = e \lim_{n\to\infty} \frac{n^s}{2^n} = 0 \end{align}

Answer 5

Ignoring the hint, let $m=\lceil s\rceil$. Then, for $n\gt m+1$, we have

$${n!n^s\over n^n}\le{n!n^m\over n^n}=\left(n\over n\right)\left(n-1\over n\right)\cdots\left(m+2\over n\right)\left(m+1\over n\right)\left(mn\over n\right)\left((m-1)n\over n\right)\cdots\left(2n\over n\right)\left(1n\over n\right)\\ \le1\cdot1\cdots1\cdot\left(m+1\over n\right)m(m-1)\cdots2\cdot1={(m+1)!\over n}$$