If $x$ is any real number then, $$\lim_{n\to \infty } \, \frac{x^2}{n!}=0$$

For part (b), pick any $x$ and let $N$ be an integer such that $N > |x|$. If $n > N$ we have

$$\begin{align}\Big|\frac{x^n}{n!}\Big| &= \frac{|x|}{1}\frac{|x|}{2}\frac{|x|}{3}\cdots \frac{|x|}{N-1}\frac{|x|}{N}\frac{|x|}{N+1}\cdots\frac{|x|}{n}\\&<\frac{|x|^{N-1}}{(N-1)!}\frac{|x|}{N}\frac{|x|}{N}\frac{|x|}{N}\cdots\frac{|x|}{N}\\&=\frac{|x|^{N-1}}{(N-1)!}\Bigg(\frac{|x|}{N}\Bigg)^{n-N+1} = K\Bigg(\frac{|x|}{N}\Bigg)^{n}\end{align}$$

Now I get stuck with the terms < $$\frac{\left| x\right| }{N}\frac{\left| x\right| }{N}\frac{\left| x\right| }{N}...\frac{\left| x\right| }{N}$$

First of all I do not see why we after we have expanded it in the first row why we set this less than, second row. I also do not see why we have more than one fraction with $N $ in the denominator on the second line.

I understand the objective to show that $0$ times a constant $K$ is equal to zero. But could someone take me through this step by step. I usually get induction but I'm just stuck with this one.

  • 1
    $\begingroup$ Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. $\endgroup$ – AlexR Jan 11 '15 at 22:21
  • $\begingroup$ Note that $\lim_{n\to\infty}\sum_{k=0}^n\frac{x^k}{k!}=e^x$ (and the series converges absolutely for all $x\in\mathbb C$) so necessarily $\lim_{k\to\infty}\frac{x^k}{k!}=0$. (I know this doesn't answer your question, but just something to keep in mind.) $\endgroup$ – Math1000 Jan 11 '15 at 22:28

Notice that to get the enequality "<" $$N + 1 > N \Rightarrow \frac{1}{N+1} < \frac{1}{N} \Rightarrow \frac{|x|}{N+1} < \frac{|x|}{N}\ \text{and} \ \ n > N \Rightarrow \frac{1}{n} < \frac{1}{N}$$

As for the second question you are taking $n > N$ so

$$1 , 2 , \ldots , N, N + 1, \ldots , n$$


One more thing $$\frac{|x|}{1}\frac{|x|}{2}\frac{|x|}{3}\cdots\frac{|x|}{N-1} = \frac{|x|^{N-1}}{(N-1)!}$$

  • $\begingroup$ Feel free to ask any questions. $\endgroup$ – Aaron Maroja Jan 11 '15 at 21:59
  • $\begingroup$ I would understand in row 2 if was as you described for the answer to my second question but it is multiple fractions with only N in the denominators. Why not ex. N-1, N, N+1 $\endgroup$ – ALEXANDER Jan 11 '15 at 22:03
  • $\begingroup$ To each $N+1, N+2,\ldots, n$ you do as I showed on the asnwer, then you'll get each term of the multiplication, since $a<c \Rightarrow ab<cb$, if $b > 0$. $\endgroup$ – Aaron Maroja Jan 11 '15 at 22:10
  • $\begingroup$ And notice that the terms $1,2, \ldots,N-2,N-1$ were already use to get $(N-1)!$ $\endgroup$ – Aaron Maroja Jan 11 '15 at 22:15
  • $\begingroup$ I understood the last comment as well as your edit, but I do not get why we multiply in the second line this with $\frac{\left| x\right| }{N}$*$\frac{\left| x\right| }{N}$.....,I dont see why N does not increase with 1 in each step. $\endgroup$ – ALEXANDER Jan 11 '15 at 22:27

It is easy to prove that, for any positive real $x$, $\lim_{n \to \infty} \frac{x^n}{n!} = 0 $.

Note that this is much stronger than $\lim_{n \to \infty} \frac{x^s}{n!} = 0 $ for any positive real $s$.

Let $r_n(x) =\frac{x^n}{n!} $ and $p(x) =\frac{x^{\lceil 2x \rceil}}{\prod\limits_{j=1}^{\lceil 2x \rceil} j} $. Then, for $n > \lceil 2x \rceil$, $$\frac{x^n}{n!} =\frac{x^n}{\prod\limits_{j=1}^{\lceil 2x \rceil} j \prod\limits_{j=\lceil 2x \rceil+1}^n j} =\frac{x^{\lceil 2x \rceil}}{\prod\limits_{j=1}^{\lceil 2x \rceil} j} \frac{x^{n-\lceil 2x \rceil}}{\prod\limits_{j=\lceil 2x \rceil+1}^n j} <p(x)\frac{x^{n-\lceil 2x \rceil}}{\lceil 2x \rceil^{n-\lceil 2x \rceil}} <p(x)\left(\frac{x}{\lceil 2x \rceil}\right)^{n-\lceil 2x \rceil} \le \frac{p(x)}{2^{n-\lceil 2x \rceil}} $$ or, for $n > 0$, $$\frac{x^{n+\lceil 2x \rceil}}{(n+\lceil 2x \rceil)!} \le \frac{p(x)}{2^{n}}. $$ Since $p(x)$ is a function only of $x$, $\frac{p(x)}{2^{n}} \to 0 $.


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