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 $n\to\infty$.

  • $\begingroup$ thanks. I have found another method under your enlightenment. $\endgroup$ – 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 $$\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}.$$


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