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How does one prove this limit? $$\lim\limits_{n \to \infty}\dfrac{5^n}{n!} = 0$$ L-Hospital would work, but only in an "intuitive" sense. (Or at least, I don't see how L-Hospital would work.) I was considering $\delta$-$\epsilon$ as well, but finding a way to make, given $\epsilon > 0$, a $K(\epsilon) > 0$ such that $\forall n > K(\epsilon)$ $$\left|\dfrac{5^n}{n!}\right| < \epsilon$$ seemed to be rather difficult.

What would you suggest? Assume no further background than an introductory real analysis background (no farther than series of functions).

Martin Sleziak
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Clarinetist
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  • I was thinking about the Taylor expression of exponential – Maman Nov 01 '14 at 20:43
  • Does that even work, since we don't have series here? – Clarinetist Nov 01 '14 at 20:44
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    For $n > 9$, write $$\frac{5^n}{n!} = \frac{5^9}{9!} \prod_{k=10}^n \frac{5}{k}.$$ – Daniel Fischer Nov 01 '14 at 20:44
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    For any $n\geq 10$, we have that $\frac{5}{n}\leq \frac{1}{2}$. Therefore $$\frac{5^{n+1}}{(n+1)!}= \left(\frac{5^n}{n!}\right)\cdot\left(\frac{5}{n+1}\right)\leq \left(\frac{5^n}{n!}\right)\cdot \left(\frac{1}{2}\right)$$ Thus, once $n\geq 10$, every term is at most half the size of the previous term. It is fairly easy to turn this into a $\delta$-$\epsilon$ proof. – curious Nov 01 '14 at 20:46
  • @StevenTaschuk - Thank you! I had a feeling that such a question MUST have been asked before. – Clarinetist Nov 01 '14 at 20:49
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    The stirling's approximation would be an other solution. But it would be cheating ;-) – Pierre Alvarez Nov 01 '14 at 20:49
  • See http://en.wikipedia.org/wiki/Factorial#Rate_of_growth_and_approximations_for_large_n for asymptotic expressions for $n!$ for large $n$. – Urgje Nov 01 '14 at 20:56

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Well, so "no farther than series of functions" means we already have power and Taylor series, so

$$\sum_{n=0}^\infty\frac{5^n}{n!}=e^5\implies\;\text{the series converges}\;\implies \lim_{n\to\infty}\frac{5^n}{n!}=0$$

Timbuc
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Then $\lim\limits_{n}\frac{5^n}{n!}\leq \lim\limits_n\frac{5^n}{10^n} =\lim\limits_n\left(\frac1{2}\right)^n = 0.$

which gives the required result by the sandwich theorem.

You just need to prove that $10^n < n!$ for all $n > N$ where you have to find $N$.

HINT: Use the representation $\frac{5^n}{n!}=\frac{5}{n}\frac{5}{n-1}\dots\frac{5}{1}$ and induction.

Trajan
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Hint: Let $a_n = \frac{5^n}{n!}$.

$$\lim_{n\to\infty}\Bigg|\frac{a_{n+1}}{a_n}\Bigg|=\lim_{n\to\infty}\frac{5^{n+1}}{(n+1)!}\frac{n!}{5^n} = \lim_{n\to\infty}\frac{5}{n+1}= 0 < 1$$

Aaron Maroja
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