About the limit $\lim_{n\to +\infty}\frac{n^k}{n!}$ for a fixed $k\in\mathbb{N}$.

Question

Given a natural number $k$ and some real number $\epsilon>0$, I have to prove that there exists a natural number $n$ such that $\frac{n^k}{n!}<\varepsilon$.

I tried to develop for $f(n+1)/f(n)$ and got a relation $1/(1+n) \cdot (1+1/n)^k$. Which for $n=k$ will give me the euler constant. But I could not go further and prove that for any $n$ there will be a limit and I could not find this upper bound...any suggestion (please)?

Answer 1

Given $k$, let $a_n=\frac{n^k}{n!}$. If we manage to prove $$ \lim_{n\to +\infty} a_n = 0 \tag{1}$$ then we're done, since that implies that for any $\epsilon>0$ there is some $N_\varepsilon$ for which $n> N_{\varepsilon}$ implies $\frac{n^k}{n!}<\varepsilon$. On the other hand, since $n!\geq\left(\frac{n}{e}\right)^n$, it follows that: $$ a_n \leq e^n n^{k-n} = e^k\left(\frac{e}{n}\right)^{n-k} \tag{2}$$ hence assuming $n>2e$ we have: $$ a_n \leq \frac{(2e)^k}{2^n} \tag{3}$$ and $(1)$ is trivial.

Answer 2

Hint: $\dfrac{f(n+1)}{f(n)}=\left( 1+\dfrac{1}{n}\right)^k\dfrac{1}{n+1}$ and so $$\lim_{n\to +\infty}\dfrac{f(n+1)}{f(n)}=0$$

Now use the following proposition (that you can prove) to prove that $\lim_{n\to +\infty}f(n)=0$:

Let $\{u_n\}$ be a sequence of nonnegative real numbers such as $\left\{\dfrac{u_{n+1}}{u_n}\right\}$ converges to $\ell\in\mathbb{R}$.

1. If $\ell <1$ then $\lim\limits_{n\to +\infty }u_n=0$
2. If $\ell >1$ then $\lim\limits_{n\to +\infty}u_n=+\infty$

Hint 2: You can also use Stirling's Formula:$$n!\sim \sqrt{2\pi n}(ne^{-1})^n$$

Answer 3

Hint: In other words you have show that $$\lim\limits_{n\to\infty}\frac{n^k}{n!}=0.$$

An easy way to do this (and your attempt with $\dfrac{f(n+1)}{f(n)}$ might be useful here) is to look at the series $\displaystyle\sum\limits_{n=1}^\infty \frac{n^k}{n!}$. If you can prove that this series is (absolutely) convergent...

Answer 4

In this answer, Bernoulli's Inequality is used to show that $\left(1+\frac1n\right)^n$ is an increasing sequence. In that same answer, it is shown, again using Bernoulli's Inequality, that $\left(1+\frac1n\right)^{n+1}$ is a decreasing sequence. Thus, since $\left(1+\frac15\right)^6\lt3$, we have that $\left(1+\frac1n\right)^n\lt3$ for all $n\ge1$.

Thus, for $n\ge\max(k,5)$, we have $$ \left.\frac{(n+1)^k}{(n+1)!}\middle/\frac{n^k}{n!}\right.=\frac{\left(1+\frac1n\right)^k}{n+1}\lt\frac 36=\frac12 $$ Therefore, $$ \frac{n^k}{n!}\le\frac{\max(k,5)^k}{\max(k,5)!}\left(\frac12\right)^{n-k} $$ Using logs, it is not too hard to compute the $n$ that would make the right side less than $\varepsilon$.