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I was working through the proof of the convergence of the Maclaurin series for certain functions, and I came across this limit, which I want to prove, but am not sure how to prove.

$\lim_{x\to\infty} \frac{a^x}{x!}=0$, $a>0, a\in N$

For this limit, techinically $x!$ should grow faster than $a^x$ for any value of $a$, but when $a$ is really big, then the limit becomes hard to prove.

Using L'Hospital is not really feasible, as an exponential function always has an exponential in its derivative and the factorial function doesn't have a derivative.

I then tried using the definition of limits: Limit is true when for all $\varepsilon>0$ there exists $m$ such that if $x>m$, then $∣f(x) ∣ = |\frac{a^x}{x!}|<\varepsilon$. But then I have no clue what value $m$ should be relative to $\varepsilon$. I'm not sure if this is the way to solve the limit.

Aarony Jamesys
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  • Use a comparison. Note that $\frac {a^x}{(2a)^x} > \frac{a^x}{x!}$ for large enough x because you are multiplying larger and larger $x$ and you can show $x! > (2a)^x$ for large enough x easily. Now if the left hand has a limit of 0 and right hand is always positive... use squeeze theorem or whatever its called – nabu1227 Apr 07 '21 at 03:38

2 Answers2

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$\frac {a^x}{x!} = \prod_{k=1}^x \frac ak$.

Now lets suppose $x$ is significantly larger than $a$. Let $m$ be an integer that is larger than $a$ but less than $x$. ($m = \lceil a \rceil$ is the perfect choice but any integer larger than $a$ will do).

The $\frac {a^x}{x!} = \prod_{k=1}^x \frac ak=$

$(\prod_{k=1}^{m} \frac ak)\cdot (\prod_{j=m+1}^x \frac aj)=\frac {a^m}{m!}\cdot (\prod_{j=m+1}^x \frac aj)$

Now $\prod_{k=1}^{m} \frac ak=\frac {a^m}{m!}$ is a set constant. Let's label it $C:=\frac {a^m}{m!}$. ANd $\frac aj < \frac am < 1$.

So $\frac {a^x}{x!} = (\prod_{k=1}^{m} \frac ak)\cdot (\prod_{j=m+1}^x \frac aj)<$

$\frac {a^m}{m!}\cdot (\prod_{j=m+1}^x \frac am)$

$C \cdot (\frac am)^{x-m}$

And as $\frac am < 1$ we have

$\lim_{x\to \infty}C \cdot (\frac am)^{x-m} = 0$

fleablood
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$$y=\frac{a^x}{x!}\implies \log(y)=x \log(a)-\log(x!)$$ Using Stirling approximation $$\log(y)= (1+\log (a)-\log (x))x-\frac{1}{2} \log \left({2 \pi x}\right)-\frac{1}{12 x}+O\left(\frac{1}{x^3}\right)$$ Since $a$ is fixed, there is $x_*=a e$ such that if $x>x_*$ which makes the leading term negative.

Claude Leibovici
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