# Convergence of Series w/ factorial

I'm working on a convergence problem that's giving me trouble. I'll list the steps I've made so far.

Given the following series determine if it is convergent or divergent: $$\sum_{n=1}^{\infty}\frac{n!\cdot x^n}{n^n}, \text{where } x > 0.$$

When I first saw this problem I thought to use the root test so I attempted to preform the following calculation: $$\lim_{n\to\infty} \sqrt[n]{\left| \frac{n!\cdot x^n}{n^n} \right|}.$$

But here is where I'm not sure how to move forward. I'm basically unsure if we can distribute the $\frac{1}{n}$ exponent to $n!$ to generate something like this ${(n!)}^{1/n} \cdot x$ as the numerator (which would go to $x$ as $n \longrightarrow \infty$).

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You had $\lim_{x\to\infty}$; I corrected that, which was probably just a slip, to $\lim_{n\to\infty}$. – Brian M. Scott Apr 30 '12 at 0:41
Not quite. The $x$ part is fine, for the rest use the Stirling approximation. Don't need quite the full approximation, just a fairly crude estimate for $\log(n!)$. – André Nicolas Apr 30 '12 at 0:45
@AndréNicolas does this mean you are suppose to replace n! completely with the stirling approximation and preform the root test normally? – Math_Illiterate Apr 30 '12 at 0:47
The $n$-the root of $n!/n^n$ approaches $1/e$. – André Nicolas Apr 30 '12 at 1:07
I've figured it out using stirlings approximation, thanks – Math_Illiterate Apr 30 '12 at 1:46

Here we can make use of Stirling's approximation, which gives us that $n!=\frac{n^n}{e^n}O(n)$, that is that $n!$ divided by $\frac{n^n}{e^n}$ is at most linear for large $n$. Thus $$\sum\limits_{n=1}^\infty \frac{n!\cdot x^n}{n^n}= \sum\limits_{n=1}^\infty \frac{n^n\cdot x^n\cdot O(n)}{e^n\cdot n^n}= \sum\limits_{n=1}^\infty \left(\frac{x}{e}\right)^nO(n)$$ which converges when $|x|<e$ by the ratio test. Using a more refined version of Stirling's, we get that $n!>c\frac{n^n}{e^n}$ for some constant $c>0$ and so $$\sum\limits_{n=1}^\infty \frac{n!\cdot x^n}{n^n}> \sum\limits_{n=1}^\infty \frac{n^n\cdot x^n\cdot c}{e^n\cdot n^n}= c\sum\limits_{n=1}^\infty \left(\frac{x}{e}\right)^n$$ which diverges when $|x|\geq e$. Thus the series converges if and only if $|x|<e$, so restricting our attention to $x>0$ we have that the series converges for $x<e$.

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Doesn't the second series converge for $x<0$ also? Plus, note the condition you want is $|x|<e$. $\color{red}{0<}x < e$ suffices, but it can also be the case that $-e < x < 0$ – Pedro Tamaroff Apr 30 '12 at 1:26
@PeterTamaroff I restricted my attention to $x>0$ b/c the OP specified as such. – Alex Becker Apr 30 '12 at 1:42
I know. However, I think it is misleading to write if and only if $x < e$, when it is actually $|x|<e$. Maybe you can say something like "Since the series converges for $|x|<e$, then $0<x<e$ works" or something of the sort. BTW, I upvoted, so don't let me down :). – Pedro Tamaroff Apr 30 '12 at 2:00

In lack of Stirlings approximation, one can use D'Alambert's criterion. We know a series converges if the ratio of its consecutive elements is eventually $<1$ in absolute value.

We have

$$a_n=\frac{{n!}}{{{n^n}}}x^n$$

Then we find

$$\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {x\frac{{{n^n}}}{{n!}}\frac{{\left( {n + 1} \right)!}}{{{{\left( {n + 1} \right)}^{n + 1}}}}} \right|$$

$$\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{n^n}}}{{{{\left( {n + 1} \right)}^n}}}x} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {{{\left( {\frac{n}{{n + 1}}} \right)}^n}x} \right|$$

$$\mathop {\lim }\limits_{n \to \infty } \frac{{\left| x \right|}}{{\left| {{{\left( {1 + \frac{1}{n}} \right)}^n}} \right|}} = \frac{{\left| x \right|}}{e}$$

Since we want the limit to be less than unity, it is manifest we need:

$$\frac{{\left| x \right|}}{e} < 1 \Rightarrow \left| x \right| < e$$

which is what Alex showed.

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