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I want to show that $\frac{x^{n}}{n!}$ goes to $0$ for $n$ going to infinity for all $x \gt 0$.

I wish to prove this using the $\varepsilon$-$\delta$ method. So for all $\varepsilon > 0$ there exist $\delta >0$ so that $\frac{x^{n}}{n!} < \varepsilon$ where $n > \delta$.

This is not a duplicate since I ask for a delta-epsilon proof.

Ole Petersen
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  • Note that $\varepsilon - \delta$ reasoning is used for functions, whereas for each $x$ you deal with a sequence, so you want to use the $\varepsilon - N_\varepsilon$ method, which you will find among the answers to the linked question. – Alex M. Dec 18 '15 at 11:51
  • It's not duplicate since I asked for a delta-epsilon proof. – Ole Petersen Dec 18 '15 at 12:27

2 Answers2

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You can use the fact that $n!\le n^n$, this means that $|x^n/n!|\le(x/n)^n$, but if $n>|2x|$ you have $|x/n|<1/2$ so $|x^n/n!| < 2^{-n}$ then.

This means that if you chosse $\delta \ge \max(|2x|, -\log_2\epsilon)$ you would be guaranteed that whenever $n>\delta$ that ($n>|2x|$ and therefore) $x^n/n! < 2^{-n} < 2^{log_2\epsilon} = \epsilon$.

skyking
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  • The last line should therefore give $\epsilon ^{-1}$ and not $\epsilon$? We know that there exist $\delta_{1} > \mid 2x\mid$ and $\delta_{2} > \log_{2} \epsilon$. How can we justify that $\delta$ even exist where $\delta = \delta_{1} = \delta_{2}$? – Ole Petersen Dec 18 '15 at 10:30
  • Should $\delta$ depend on $x$? :) – Hosein Rahnama Dec 18 '15 at 10:36
  • @OlePetersen Well, observed. I changed the sign so it's correct. Such a $\delta$ exists because there's always numbers larger than two arbitrarily chosen numbers. – skyking Dec 18 '15 at 10:49
  • @H.R. Yes, that's my interpretation of the statement to be proved. That is for all $x>0$ it's true that $\lim x^n/n! = 0$. – skyking Dec 18 '15 at 10:50
  • I understand the proof now. So for notation purpose we can actually choose $\delta = \max (\delta_{1}, \delta_{2})$. – Ole Petersen Dec 18 '15 at 10:53
  • Ah, I just got a stupid confusion! :) That's OK! :) – Hosein Rahnama Dec 18 '15 at 10:57
  • @OlePetersen As that maybe suit you better, I've change the formulation there. – skyking Dec 18 '15 at 11:00
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Note that $x>0$ is given once and for all. Put $N:=\lceil 2x\rceil\geq 2x$. Then one has $$0<{x^n\over n!}\leq {x^N\over N!}\left({1\over2}\right)^{n-N}={M\over 2^n}<{M\over n}\qquad(n\geq N)\ .\tag{1}$$ Here $M:={2^N x^N\over N!}$ is some fixed large number, and we have used the fundamental fact that $2^n>n$ for all $n\geq0$.

Now let an $\epsilon>0$ be given. If $$n>n_0:=\max\left\{N,{M\over\epsilon}\right\}$$ then $(1)$ implies $\ {\displaystyle 0<{x^n\over n!}<\epsilon}$.

Christian Blatter
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