It’s been a while since my last update. A few weeks ago, my colleague proposed a question regarding the limit of $\frac{ n }{\sqrt[n]{n!}}$ when teaching calculus. I found it interesting to make use of the lemma below to derive some limts:

Lemma. Suppose $\lim\limits_{n\rightarrow \infty} a_n = a$ and $a_n>0,~\forall n$. Then $\lim\limits_{n\rightarrow \infty} \sqrt[n]{a_1\cdots a_n} = a.$

  • Example 1. $\lim\limits_{n\rightarrow \infty} \frac{n+1}{n} = 1.$ By the Lemma, we have:
\[\lim\limits_{n\rightarrow \infty} \sqrt[n]{\frac{2}{1}\frac{3}{2}\cdots\frac{n+1}{n}} = \lim\limits_{n\rightarrow \infty} \sqrt[n]{n+1} = 1.\]
  • Example 2. $\lim\limits_{n\rightarrow \infty} \frac{1}{n} = 0.$ By the Lemma, we have:
\[\lim\limits_{n\rightarrow \infty} \frac{1}{\sqrt[n]{n!}} = 0 ~~\Leftrightarrow~~ \lim\limits_{n\rightarrow \infty} \sqrt[n]{n!} = \infty.\]
  • Example 3. $\lim\limits_{n\rightarrow \infty} \left(1+\frac{1}{n}\right)^n = e.$ By the Lemma, we have:
\[\begin{align} e &= \lim\limits_{n\rightarrow \infty} \left(1+\frac{1}{n}\right)^n\\ &= \lim\limits_{n\rightarrow \infty} \sqrt[n]{\left(\frac{2}{1}\right)\left(\frac{3}{2}\right)^2\cdots\left(\frac{n+1}{n}\right)^n}\\ &= \lim\limits_{n\rightarrow \infty} \sqrt[n]{\frac{(n+1)^n}{n!}}\\ &= \lim\limits_{n\rightarrow \infty} \frac{ n+1 }{\sqrt[n]{n!}} = \lim\limits_{n\rightarrow \infty} \frac{ n }{\sqrt[n]{n!}}, \end{align}\]

where the last equality follows from Example 2. Equivalently, we also have

\[\lim\limits_{n\rightarrow \infty} \frac{\sqrt[n]{n!}}{ n } = \frac{1}{e}.\]

Note that this important limit can also be obtained as follows:

\[\begin{align} \lim\limits_{n\rightarrow \infty} \frac{\sqrt[n]{n!}}{ n } &= e^{\log \lim\limits_{n\rightarrow \infty} \frac{\sqrt[n]{n!}}{ n } }\\ & = e^{\lim\limits_{n\rightarrow \infty} \log \sqrt[n]{\frac{n!}{n^n}} } = e^{\lim\limits_{n\rightarrow \infty} \frac{\log \frac{1\cdots n}{n \cdots n}}{n}} \\ & = e^{\lim\limits_{n\rightarrow \infty} \frac{1}{n}\sum\limits_{i=1}^n \log \left(\frac{i}{n}\right) }\\ & = e^{\int_0^1 \log x~dx} = \frac{1}{e}. \end{align}\]