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$\lim\limits_{n\to\infty} \frac{a^n}{n!}=0$

how to prove this one since L'Hospital Theorem cannot be used with factorial?

José Carlos Santos
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RexSolus
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    note that for $n$ big enough $a<n$ so the terms $\frac an<1$. Multiply enough of them to get as small as you need. – Gregory Grant Jul 25 '17 at 10:39
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    L'Hospital is not the alpha and omega of limits computation. Nor does it prepare your coffee. In fact it should be avoided as much as possible: when it works, Taylor's formula at order $1$ works as well, without its drawbacks. – Bernard Jul 25 '17 at 10:40
  • Use the ratio test. – Bernard Jul 25 '17 at 10:41
  • @Bernard superb comment, made me chuckle :D – AlvinL Jul 25 '17 at 10:41
  • Regards @RexSolus , you may notice that $\frac{a^{n}}{n^{n}} < \frac{a^{n}}{n!} $. The left side converges to 0. Now you may prove that your sequence is decreasing for $a<n+1$. – Redsbefall Jul 25 '17 at 10:46
  • @Bernard what "drawbacks" does a first order Taylor expansion avoid? – Ben Grossmann Jul 25 '17 at 10:53
  • One often for gets to check all the hypotheses in L'Hospital's rule. In particular, one must ensure that, in a small neighbourhood of $a$, $g(x)\ne 0$ – except at $a$ itself, of course. – Bernard Jul 25 '17 at 10:57

6 Answers6

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Hint. Let $N\geq |a|$, then for $n>N$, we have that \begin{align*} 0\leq \frac{|a|^n}{n!}&=\frac{|a|}{n}\cdot \frac{|a|^{n-N-1}}{(n-1)\cdots (N+1)}\cdot \frac{|a|^{N}}{N!}\\ &\leq \frac{|a|}{n}\cdot \left(\frac{|a|}{N+1}\right)^{n-N-1}\cdot \frac{|a|^{N}}{N!}\\ &\leq \frac{|a|}{n}\cdot \frac{|a|^{N}}{N!} \end{align*} then apply the Squeeze Theorem as $n\to \infty$.

Robert Z
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  • So, if I let n be near infinity for the right side ab(a)/n will be zero and therefore the term be zero, which means that my limit is zero.( Squeeze Theorem ). Am I right with that? if I right, I wonder that why the symbol is ≤ not = – RexSolus Jul 25 '17 at 10:56
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    @RexSolus Yes, taking the limit both sides we get $0\leq \lim_{n\to\infty} \frac{|a|^n}{n!}\leq 0$, which implies equality. – Robert Z Jul 25 '17 at 10:59
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Stirling: $n! \sim (\frac{n}{e})^n \sqrt{2\pi n}$

$\frac{a^n}{n!} \sim \frac{a^n e^n}{n^n\sqrt{2\pi n}} = (\frac{a\times e}{n})^n \frac{1}{\sqrt{2\pi n}} = \exp(n \ln(\frac{a\times e}{n})) \frac{1}{\sqrt{2 \pi n}} \rightarrow 0$

Wylex
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$$\frac{a^n}{n!} = \frac{a^k}{k!}\cdot \frac{a^{n-k}}{(k+1)\cdot(k+2)\cdots n!}\leq C\cdot\frac{a^{n-k}}{(a+1)^{n-k}}$$

for some constant $c$ and a large enough value of $k$ (i.e., a value of $k$ such that $k>a$).

5xum
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One can also show by D'Alembert's criterion criterion that the series $$\sum_{n=1}^{\infty} \frac{a^n}{n!}$$ converges, so the limit of the summand must be zero.

(This is rather a funny fact than a real solution, since the technique used in the proof of the criterion can be used directly to prove that the limit is zero).

Adayah
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If we apply the ratio test, we find that $$ \lim_{n \to \infty}\frac{a^{n+1}}{(n+1)!}\cdot \frac{n!}{a^n} = \lim_{n \to \infty} \frac{a}{n+1} = 0 $$ Typically, we use this to deduce that $\sum \frac{a^n}{n!}$ converges. However, it is certainly enough to deduce that $\frac{a^n}{n!} \to 0$, since this is a requirement for the above convergence.

Ben Grossmann
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$$0<\left| \frac { { a }^{ n } }{ n! } \right| =\frac { \left| a \right| }{ 1 } \frac { \left| a \right| }{ 2 } \frac { \left| a \right| }{ 3 } ...\frac { \left| a \right| }{ m } \quad \frac { \left| a \right| }{ (m+1) } ...\frac { \left| a \right| }{ n } \le \frac { \left| { a }^{ m } \right| }{ m! } { \left( \frac { \left| a \right| }{ m+1 } \right) }^{ n-m }<\varepsilon \\ $$ for every $\forall \varepsilon >0$ and $m+1 >\left|a\right|$

haqnatural
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