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$$\forall \epsilon > 0, \exists N \in \mathbb{N} \text{ such that } n > N \Rightarrow \left| \frac{1}{a_n} \right| < \epsilon$$

If this is true, we know that the sequence ${a_n}$ diverges as $ n \rightarrow \infty$

I got stuck in this proof as I want to show that:

$$\left|\frac{20^n}{n!}\right| < \epsilon$$

by a line of inequalities, I can state that $$\frac{20^n}{n!} < 20^n < \epsilon \Rightarrow -n > -\frac{\log(\epsilon)}{\log(20)}$$

Using Python this result gives nonsensical answers, it's obvious it would be from the equation as it's not true for all $N$... Does anyone have any advice on how to go about fixing it? Thank you!

Furthermore, I cannot use any tests it has to be a proof by definition.

Ma.Te.Pa
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  • Please use the appropriate commands "\log" and "\lim" for logarithms and limits. – Digitallis Sep 26 '23 at 16:42
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    Hint: For $n\geq 39,$ $a_{n+1}\geq 2a_n.$ – Thomas Andrews Sep 26 '23 at 16:46
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    You're certainly not going to show that $20^n<\epsilon$ for arbitrarily large $n$. Also if you did somehow show that (which you can't because it's false), you would be done. Finally, if you were to continue reasoning after that, you would not conclude $n > -\frac{\log(\epsilon)}{\log(20)}$, you would conclude $n < \frac{\log(\epsilon)}{\log(20)}$, or if for some strange reason you wanted to reverse the inequality you would have $-n>-\frac{\log(\epsilon)}{\log(20)}$. – M W Sep 26 '23 at 16:52
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    Generally you need to be more careful not to throw too much away when you develop an inequality. Like here, it is true that if you replace $n!$ with $1$ then you get something bigger, but you also now have something that doesn't go to zero anymore, so your observation is true but useless. Instead, somehow you want to replace $n!$ with something smaller (and simpler) that is still way bigger than $20^n$ if $n$ is big enough. Since $n!$ is a product of a bunch of stuff, you can consider replacing some/all of the factors in that product by smaller factors. – Ian Sep 26 '23 at 16:58
  • @ThomasAndrews Given this hint I reched an answer where $N = max{39, \frac{20}{\epsilon}}$ Though I am not sure how to reach the hint without using programming. – Ma.Te.Pa Sep 26 '23 at 17:04
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    @TeodorasPaura $a_{n+1}=\frac{n+1}{20}a_n$ makes it pretty obviously without programming. – Thomas Andrews Sep 26 '23 at 17:12
  • And $N=\max(39,20/\epsilon)$ won't work in general. – Thomas Andrews Sep 26 '23 at 17:14
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    @TeodorasPaura, your question is a duplicate. Here is an answer. – Angelo Sep 26 '23 at 18:21
  • @Angelo The question presented uses a definition rather than intuitive or other advanced methods in your linked question. – Ma.Te.Pa Sep 26 '23 at 18:42
  • Does this (more elementary) answer your question? prove $\lim_{n\to\infty}{\frac{a^n}{n!}}=0$ See also here or there. All found using Approach$0$, but there are probably more in that link. – Anne Bauval Sep 26 '23 at 21:46
  • @AnneBauval Thank you, yes. – Ma.Te.Pa Sep 26 '23 at 21:50

1 Answers1

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You can use $a_{n+1}=\frac{n+1}{20}a_n,$ and thus, for $n\geq 40,$ $a_{n+1}> 2a_n,$ for $n\geq 40.$

Then $a_n>2^{n-40}a_{40}$ for $n> 40.$ To get $a_n>\frac1\epsilon,$ you need $$n-40> \log_2\frac1{a_{40}\epsilon}$$

So you can take $$N=40+\max\left(0,-\log_2(a_{40}\epsilon)\right).$$

You can get a trivial bound $a_{40}>\frac1{20^{40}}>\frac1{32^{40}},$ so $-\log_2 a_{40}\leq 200.$ You can do much better than this with a better estimate of $a_{40},$ but you don't need it. Then you can use:

$$N=40+\max(0,200-\log_2\epsilon).$$

Thomas Andrews
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