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Find the convergence radius of the serie

$$\sum \frac{n^n}{n!}x^n $$

and analyze the absolute convergence and/or uniform.

What I've done:

  • It is easy to show that the radius of convergence of this series is $R=\frac{1}{e}$. Then, the series convergence absolutely and uniformly on the interval$\; \left(-\frac{1}{e}, \frac{1}{e} \right)$

  • Analyzing the convergence at $x=\frac{1}{e}$ (see here), the serie does not comverges.

  • Analyzing the convergence at $x=\frac{-1}{e},$ the Dirichlet test says that the serie converges, if I didnt do anything wrong, using the same factorial aproximation we've seen before

I have some questions about what I've done and the difference between absolute and uniform convergence.

  1. Since the uniform convergence talks about series of functions, I think that does not makes sence to "analyze the uniform convergence" st the interval extremes. Is that correct?But it does makes sence to talk about uniform convergence at $[-\frac{1}{e}, \frac{1}{e})$! I am confuse about this part.

  2. I think the final answer should be that the series converges absolutely and uniformly at the interval of convergence and converges at the interval of convergence and for $x=\frac{1}{e},$ if I didnt do anything wrong.

I hope my questions are clear. Thanks for your help!

Community
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Giiovanna
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  • You have to be a little careful about applying the Dirichlet test here - you need very explicit bounds on the magnitude of $\dfrac{n^n}{n!e^n}$, and bounds that are good enough to show that the terms are strictly decreasing in magnitude - just saying that they're of 'size $\dfrac1{\sqrt{n}}$' isn't quite good enough. The bounds mentioned in your other question should be enough to prove the strict inequality, but you do need to do that before you can use Dirichlet. – Steven Stadnicki Sep 25 '14 at 22:35
  • (As an example of what can go wrong, imagine the sequence $a_{2n} =\dfrac1{\sqrt{2n}}$ and $a_{2n+1} = \dfrac5{\sqrt{2n+1}}$. Then $\lim a_n = 0$, but you can't use Dirichlet to say that $\sum (-1)^na_n$ converges.) – Steven Stadnicki Sep 25 '14 at 22:38
  • But my function is monotonic, isn'T it? – Giiovanna Sep 25 '14 at 22:39
  • Not until you've proved it's monotonic. :-) – Steven Stadnicki Sep 25 '14 at 22:40
  • Well, if I didnt do any wrong calculation, I proved that it is. – Giiovanna Sep 25 '14 at 22:41
  • $\frac{a_{n+1}}{a_n} = (1+\frac{1}{n})^n \frac{1}{e} < 1$, correct? – Giiovanna Sep 25 '14 at 22:44
  • Ahhh, of course - that's a much clearer way of doing it than using any asymptotic estimates, to be sure. That just hadn't gotten mentioned in your question so I figured it was worth calling out explicitly. – Steven Stadnicki Sep 25 '14 at 22:45
  • In this case, the Dirichlet test applies! But what aboutthe uniform convergence at [-1/e, 1/e)? Should it be considered (analyzed)? I am really confused about where should I test it (clearly, not out of [-1/e,1/e]) – Giiovanna Sep 25 '14 at 22:47
  • That, sadly, I can't really help you with - issues of uniformity are not my strong point! – Steven Stadnicki Sep 25 '14 at 22:53

1 Answers1

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The convergence radius $r$ of $\sum a_nx^n$ is given by: $$ r^{-1}=\limsup_{n\to\infty}|a_n|^{1/n} $$ By Stirling approximation: $$ e=\lim_{n\to\infty}\left(\frac{e^n}{e\sqrt{ n}}\right)^{1/n}\leq\limsup_{n\to\infty}\left(\frac{n^n}{n!}\right)^{1/n}\leq\lim_{n\to\infty}\left(\frac{e^n}{\sqrt{2\pi n}}\right)^{1/n}=e $$ Therefore $r=1/e$. The series converges absolutely for $|x|<1/e$ and converges uniformly on compact subset of $]-1/e,1/e[$. Since the series converges, as the PO said, for $-1/e$, it also converges uniformly on any compact subset of $[-1/e,1/e[$ (just pick the larger $N$ of the two as in $n>N\Rightarrow |f_n(x)-f(x)|<\varepsilon, \forall x\in]-1/e,1/e[$ and in $x=-1/e$).

The series is not uniformly convergent on $[-1/e,1/e[$, for otherwise $|\sum_{k=n}^mf_k(x)|\leq\varepsilon$ for $f_k(x)= \frac{k^k}{k!}x^k,\forall n,m>N, \forall x\in [-1/e,1/e[$ (uniformly cauchy=uniformly convergent on $\mathbb R$); as $x\to 1/e$, since $f_k(x)$ are all continuous, $|\sum_{k=n}^mf_k(1/e)|\leq\varepsilon$. We arrive at the conclusion that the series converges for $x=1/e$, which is a contradiction.

In my intuitive understanding, since the series blows up at $x=1/e$, it cannot have a uniform convergence as $x$ approaches $1/e$.

Troy Woo
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  • Ok, but if the series converges for $x=-\frac{1}{e}$, why the series doesn not converges uniformly on the interval $[-1/e, 1/e[$? – Giiovanna Sep 25 '14 at 22:31
  • @Giiovanna You are right, it should be uniform on a compact subset of $[-1/e,1/e[$. – Troy Woo Sep 25 '14 at 22:52
  • I dont see why your proof for the not uniformly convergence at [-1/e, 1/e) does not implies that the series also does not converges uniformly on (-1/e, 1/e) – Giiovanna Sep 26 '14 at 10:34
  • @Giiovanna Whatever you mean (I don't get you at the moment, you said it true, but I am addressing your original question, and your comment is not making sense to me), its not going to happen for the $\dots,1/e[$ part. The biggest subset you can have uniform convergence is $[-1/e,1/e-\varepsilon]$ for arbitrary small $\varepsilon>0$, but no matter how small $\varepsilon>0$ is, it is not the same as $[-1/e,1/e[$. – Troy Woo Sep 26 '14 at 11:29