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I want to check the pointwise and uniform convergence of the below power series.

  • $\displaystyle{\sum_{n=1}^{+\infty}\frac{n!}{n^n}x^n}$

We have that \begin{equation*}\sqrt[n]{\frac{n!}{n^n}}=\frac{\sqrt[n]{n!}}{n}\end{equation*}

How do we calculate that limit?

Arctic Char
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Mary Star
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2 Answers2

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From the more general result presented here

we have that

$$\frac{a_{n+1}}{a_n}\to L \implies \sqrt[n]{a_n} \to L$$

For the application see the related

user
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  • Ok, so we get that the pointwise converegnce is on $(-e,e)$. To check now the boundaries... How can we calculate the limit of $\frac{n!}{n^n}e^n$ ? I have done the following: $$\lim_{n\rightarrow +\infty}\frac{n!}{n^n}e^n=\lim_{n\rightarrow +\infty}\frac{e^n}{n^n}\cdot n!=\lim_{n\rightarrow +\infty}\left (\frac{e}{n}\right )^n\cdot n!$$ does this help us? How can we continue? Do we maybe apply the Taylor expansion of $e^x$ ? – Mary Star Aug 24 '20 at 23:55
  • @MaryStar Refer to this OP for that. – user Aug 24 '20 at 23:58
  • Thank you!! :-) – Mary Star Aug 26 '20 at 21:06
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    @MaryStar You are very welcome! Bye – user Aug 26 '20 at 21:14
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Let's use the ration test to check the convergence. One has $$\frac{\left( \frac{(n+1)!}{(n+1)^{n+1}}\right)}{\left( \frac{n!}{n^n}\right)} = \frac{n^n}{(n+1)^n} = \left( 1+ \frac{1}{n}\right)^{-n}$$

which tends to $e^{-1}$. So the radius of convergence is $e$. So the series converges for all $|x|<e$, and diverges for all $|x|>e$.

Now let's study what happened for $x=-e$ and $x=e$. One has, by Stirling formula, $$\frac{n!}{n^n}e^n \sim \sqrt{2\pi n}$$

So $$\lim_{n \rightarrow +\infty} \frac{n!}{n^n}e^n = +\infty$$

so both series $$\sum \frac{n!}{n^n}e^n \quad \text{ and } \quad \sum \frac{n!}{n^n}(-e)^n$$

diverge.

TheSilverDoe
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