How to obtain the lower bound of $\dfrac{n}{\sqrt[n]{n!}}$ by Taylor's series?

We want to prove $$\lim_{n \to \infty}a_n=\lim_{n \to \infty}\frac{n}{\sqrt[n]{n!}}=e.$$

I have some solutions for this, but I want to find another method applying the squeeze theorem. Thus, a natrual thought is to find the upper bound and the lower bound of $$a_n$$.

Notice that $$e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots.$$ If we substitute $$x$$ for $$n$$, then $$e^n=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots>\frac{n^n}{n!}.\tag1$$ Thus, we obtain $$e>\frac{n}{\sqrt[n]{n!}},$$ which shows $$e$$ is a upper bound of $$a_n$$.

But how to obtain the lower bound by $$(1)$$? Say it again. I have other methods to deal with it. I just wonder whether there is some method depending on $$(1)$$ only or not.

Yes, it can be done this way. Write again $$e^n=\displaystyle\sum_{k=0}^{\infty}t_k$$ with $$t_k=\dfrac{n^k}{k!}$$. Clearly, $$t_k\leqslant t_n$$ for all $$k$$.

Further, for $$k>2n$$ we have $$t_k=t_{2n}\dfrac{n^{k-2n}}{(2n+1)\cdot\ldots\cdot k}< 2^{2n-k}t_{2n}\leqslant 2^{2n-k}t_n$$ and thus $$e^n<(2n+1)t_n+t_n\sum_{k=2n+1}^{\infty}2^{2n-k}=(2n+2)t_n,$$ which gives you $$a_n>e(2n+2)^{-1/n}\to e$$ when $$n\to\infty$$.

• thanks. I have found another method under your enlightenment. – mengdie1982 Oct 24 '18 at 5:33

\begin{align*} e^n&=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots\\ &<(n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)(n+2)}+\cdots\right]\\ &< (n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)^2}+\cdots\right]\\ &=(n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot n\\ &=(2n+1)\cdot \frac{n^n}{n!}. \end{align*} Thus $$\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}.$$