4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $n\to\infty$.

$\endgroup$
  • $\begingroup$ thanks. I have found another method under your enlightenment. $\endgroup$ – mengdie1982 Oct 24 '18 at 5:33
1
$\begingroup$

\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 $$\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.