1
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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

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

Edit:

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $.

$\endgroup$

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.