# Prove $\lim_{n\to\infty} \frac{n^x}{n!}=0$.

I've been having trouble using the definition of a limit to prove limits, and at the moment I am trying to prove that $$\lim_{n\to\infty} \frac{n^x}{n!}=0$$

for all $x$ which are elements of natural numbers. I'm able to start the usual setup, namely let $0<\epsilon$ and attempt to obtain $\left\lvert\dfrac {n^x}{n!}\right\rvert <\epsilon$. I don't really feel like this is correct, and I have absolutely no idea how to go about proving this. Any help at all would be very much appreciated!

• Why wouldn't the usual setup be correct? But it counts asabsolutely nothing in regards to showing effort, so: What are your own thoughts? – Henrik Oct 28 '17 at 20:36
• If $x$ makes it complicated, try to prove, for example, $x=7$ and generalize it for any $x$. – ThePortakal Oct 28 '17 at 20:38
• Divide two consecutive terms from $a_n=\frac{n^x}{n!}$: $$\frac{n^x}{n!}\frac{(n+1)!}{(n+1)^x}=\frac{n+1}{((1+1/n)^n)^{x/n}}$$ The denominator tends to $e^{0}=1$ and the numerator tends to $+\infty$. – olsen5 Oct 28 '17 at 20:42
• possible duplicate use the Stirling formula – Guy Fsone Oct 28 '17 at 21:10

If $x \le 0$, then there is really nothing to show. Otherwise, if $x > 0$, let $m$ be a positive integer greater than $x$.

Then, \begin{align*} \frac{n^x}{n!} \le \frac{n^m}{n!} &= \frac{n^m}{n\cdot(n-1)\dotsb(n-m+1)}\cdot\frac{1}{(n-m)!}\\ &= \frac{n^m}{n^m\big[1\cdot\big(1-\frac{1}{n}\big)\dotsb\big(1-\frac{m+1}{n}\big)\big]}\cdot\frac{1}{(n-m)!} \\ &=\frac{1}{1\cdot\big(1-\frac{1}{n}\big)\dotsb\big(1-\frac{m+1}{n}\big)}\cdot\frac{1}{(n-m)!}. \end{align*} The first term goes to $1$ as $n\to\infty$ since each of the finitely many factors in the denominator goes to $1$ as $n\to\infty$: $$a_n=\frac{1}{1\cdot\big(1-\frac{1}{n}\big)\dotsb\big(1-\frac{m+1}{n}\big)}\to1\quad\text{as n\to\infty}.$$ Since $m$ is fixed, the second term goes to $0$ as $n\to\infty$: $$b_n = \frac{1}{(n-m)!} \to 0\quad\text{as n\to\infty.}$$ Putting it all together, $$0\le\lim_{n\to\infty}\frac{n^x}{n!} \le \lim_{n\to\infty}a_nb_n = \lim_{n\to\infty}a_n\cdot\lim_{n\to\infty}b_n = 1\cdot 0 = 0,$$ so that these inequalities must in fact be equalities, and thereby we obtain the desired conclusion.

$n!=\overbrace{1\times 2\times 3\times 4}^{>2^4}\times\overbrace{5}^{>2}\times\overbrace{6}^{>2}\times\cdots\times\overbrace{n}^{>2}\ge2^4\times2^{n-4}\ge2^n$

Or you can prove it by induction: Prove the inequality $n! \geq 2^n$ by induction

Thus $0\le\dfrac{n^x}{n!}\le\dfrac{n^x}{2^n}=\exp\big(-n\ln(2)\big(1-\frac x{\ln(2)}\underbrace{\frac{\ln(n)}n}_{\to 0}\big)\big)\to 0$

Let $\varepsilon>0$. Considering $k=\lceil x\rceil$ (the least integer greater than or equal to $x$), we have $$\frac{n^x}{n!}\le\frac{n^k}{n!}$$ Let's prove that, from $n>2k$, we have $$\frac{n^k}{n!}<\frac{2^{k+1}}{n}$$ Indeed $$n!= \underbrace{n(n-1)\dotsm(n-k)}_{\text{k+1 factors >n/2}}\,(n-k-1)\dotsm 3\cdot2\cdot 1>\left(\frac{n}{2}\right)^{k+1}$$ Now, choose $N>\max\{2k,2^{k+1}/\varepsilon\}$; then, for $n>N$, we have $$\frac{n^x}{n!}\le\frac{n^k}{n!}<\frac{2^{k+1}}{n}< \frac{2^{k+1}}{N}<\varepsilon$$

We prove for $x=4$ and generalize it from there.

Let $\varepsilon >0$. By Archimedean property, there exists $N$ such that $\frac 1N < \frac \varepsilon 8$ and $\frac{N}{N-3}\leq 2$.

Now let $n-4 \geq N$.

$$\left| \frac{n^4}{n!}\right| = \left| \frac{1}{(n-4)!} \cdot \frac{n}{n-3} \cdot \frac{n}{n-2} \cdot \frac{n}{n-1} \cdot \frac{n}{n} \right| \leq \left| \frac{1}{N!} \right| \cdot 2 \cdot 2 \cdot 2 \cdot 1 \leq \frac 8N<\varepsilon$$

Now if $x<4$, since $n^x < n^4$, the above proof also works. If $x>4,$ then a similar argument can be made for $\lceil x \rceil$ instead of $4$.

we have following theorem:

if $<a_n>$ is a sequence then if $\lim_{n \to \infty} \frac {a_{n+1}}{a_n} = 0$ then $<a_n> \rightarrow 0$

Using the above theorem

$\lim_{n \to \infty} \frac {a_{n+1}}{an} = \lim_{n \to \infty} \frac {1}{n+1} (1+\frac{1}{n})^x = 0 \implies$ given limit converges to zero