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Prove that limit of sequence $x_n = {a^n \over n!}$ is $\lim_{n \to \infty} x_n = 0$

Let $\epsilon > 0$ So we need to prove: $|x_n - a| < \epsilon$

$$\implies\left|{a^n \over n!}- 0\right| < \epsilon$$ $$\implies{a^n \over n!} < \epsilon$$ $$\implies n\log{a} - \log n!< \log\epsilon$$

I don't know how can I isolate $n$ in this inequality. Please provide hints, so that I can work the answer on my own. Thanks.

EDIT:- I think my question is not duplicate because, the answers here does not contain $\epsilon$ proof.

Community
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  • @AlexisOlson That question does not contain an epsilon proof . i am asking for epsilon proof. –  Oct 15 '16 at 06:41
  • From the estimates in the answers to that question you can easily construct an epsilon proof. – Hans Lundmark Oct 15 '16 at 10:17
  • @HansLundmark I have no problem in my question getting closed, since i got the answer to my question. But i don't think that question has the answer to my question. –  Oct 15 '16 at 12:10

3 Answers3

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Let $b\in \Bbb N$ such that $b>|a|$, Then
$$\frac{a^n}{n!}<\frac{b^n}{n!}$$ For $n\ge b$, we have $$|\frac{a^n}{n!}|<|\frac{b^n}{n!}|=\frac{b}{n}\frac{b}{n-1}...\frac{b}{b+1}(\frac{b}{b}\frac{b}{b-1}...\frac{b}{1})\le\frac{b}{n}(\frac{b}{b}\frac{b}{b-1}...\frac{b}{1})$$

Let $\epsilon>0. $ For sufficiently large $N\in \Bbb N$, we have
$$\frac{1}{n}<(\frac{b}{b}\frac{b}{b-1}...\frac{b}{1})^{-1}*\frac{\epsilon}{b}\text{ , for all }n\ge N$$ $$\Rightarrow|\frac{a^n}{n!}|<\epsilon\text{ , for all }n\ge N$$

Nick
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  • Ok, but you did the question for me. :). I have a doubt though, you say $n \ge b$, what if $n \lt b$ we did not prove for that part ? –  Oct 15 '16 at 06:39
  • We do not have to consider the situation of the first $b$ terms of $x_n$ because what we care is where does $x_n$ tend to as $n \to \infty$ – Nick Oct 15 '16 at 06:43
  • Ahh, right. thanks. You cleared many of my doubts. –  Oct 15 '16 at 06:48
  • It would be easier if you started with $b>2a$, then $x_n<\frac{a^b}{b!}2^{-(n-b)}$. – Lutz Lehmann Oct 15 '16 at 09:42
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Let \begin{align} c_n = \frac{a^n}{n!} \end{align} then observe \begin{align} \frac{c_{n+1}}{c_n} = \frac{a^{n+1}}{(n+1)!} \frac{n!}{a^n} = \frac{a}{n+1}<\epsilon \end{align} when $n$ is sufficiently large, i.e. there exists $N$ such that for all $n>N$, we have \begin{align} c_{n+1}<\epsilon c_n. \end{align} In particular, it follows \begin{align} c_{N+k}< \epsilon c_{N+k-1}<\epsilon^k c_N \rightarrow 0 \end{align} as $k\rightarrow \infty$.

Jacky Chong
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An alternative approach, not among the answers at the linked duplicate (and not an $\epsilon-\delta$ proof, but still useful for others):

Though most texts and courses assign your problem before they introduce the ratio test for convergence of series, it is perfectly valid to deduce that your limit is zero by using the ratio test.

Define $x_n=\frac{a^n}{n!}$. Then $x_{n+1}/x_n=\frac{a}{n+1}$, which obviously tends to $0$ as $n\to\infty$. By the ratio test, the series $\sum_{n=1}^\infty x_n$ converges. Hence $\lim_{n\to\infty}x_n=0$.

symplectomorphic
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  • Ok, does that mean if ratio of consecutive terms tends to a limit $x$ then the sequence will tend to $x$. right ? –  Oct 15 '16 at 06:50
  • @A---B nope, ratio test is the test which concludes whether series converges or not.. since the series converges, it is necessary for $\lim_n x_n$ to be $0$ it may sound strange if you still haven't learnt series – windircurse Oct 15 '16 at 06:59
  • Alright, so if the given $\lim_n x_n$ was $3$ in my question, then by ratio test, you would have concluded sequence converges which implies that $\lim_n x_n$ is $3$. right ? –  Oct 15 '16 at 07:04
  • @A---B: no. the ratio test says that if $\lim |\frac{x_{n+1}}{x_n}|<1$, then $\sum_{n=1}^\infty |x_n|$ converges, from which it follows that $\lim |x_n|=0$, which in turn implies $\lim x_n=0$. – symplectomorphic Oct 15 '16 at 07:15