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Exercise 4:10 in John D'Angelo's text is to find the radius of convergence for :

A) $\sum_{n=1}^\infty \frac{z^n n^n}{n!}$ and

B) $\sum_{n=1}^\infty z^{n!}$

I got half of an answer for A) which I wanted to check and I got totally stuck on B). Thanks for the help.

I know from the Theorem in the section that $\frac{1}{R} = \limsup |a_n|^{\frac{1}{n}}$ where $R$ is the radius of convergence. So,

for A:

$$\frac{1}{R} = \frac{n}{n!^{\frac{1}{n}}}\text{ so }\frac{1}{R} = \limsup \frac{n!^\left({\frac{1}{n}}\right)}{n}$$ which I think is $0$ but I'm not positive. I doubt this is correct because that would mean that the radius of convergence is $\infty$ which seems wrong.

for B:

$z^{n!} = z^{(n \times(n-1)!)}$ but I don't know how to eliminate the $(n-1)!$ so I can just have a $z^{n}$ so that I can use the theorem above regarding $R$.

Thanks again. Oh, I know I asked this before but if anyone knows of a solution manual to this text, I'd appreciate it. I'm not a student so not trying to cheat on the homework but rather just trying to understand the basics.

copper.hat
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mark leeds
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3 Answers3

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For A), try the ratio test instead: ${1 \over R} = \lim_n | {a_{n+1} \over a_n } | = \lim_n { (n+1)^{n+1} \over (n+1)!} { n! \over n^n } = \lim_n (1+ {1 \over n})^n = e$.

For B), it might be easier to compute $\limsup_n \sqrt[n]{|z|^{n!}} = \limsup_n |z|^{(n-1)!}$. For $|z|<1$ this is zero, for $|z|>1$ this is infinite.

copper.hat
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Alternative proof for (B): as $$\sum_{n=1}^N |z|^{n!}\le\sum_{n=1}^{N!} |z|^n,$$ the radius of convergence is $\ge 1$. But $$\sum_{n=1}^\infty 1^{n!}$$ diverges, so the radius is $=1$.

Martín-Blas Pérez Pinilla
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  • I'm printing out and will study answers and see if I follow. If I do, I will check off. Otherwise, I'll explain what I don't understand. thanks to all. – mark leeds Aug 20 '14 at 16:22
  • @copper: I understand what you and rehband both did for A). It's clear now. For B), I think you are applying the same formula to the series terms themselves when I thought the theorem applied to the $a_n$ coefficient multiplying the $z^{n}$ term. Could one of you clarify what the logic is there ? Thanks. – mark leeds Aug 20 '14 at 16:38
  • @RasmusE, $\sum z^n$ converges absolutely for $|z|=1-\epsilon$ $\implies$ $\sum z^{n!}$ converges absolutely for $|z|=1-\epsilon$, so the radius of convergence of $\sum z^{n!}$ is at least 1. – Martín-Blas Pérez Pinilla Aug 20 '14 at 16:49
  • @markleeds: Whenever the exponent of $z$ isn't simply a nice $n$ (here it is $n!$ for example), you can just apply the root test to the whole expression. When you do that, take the limit as $n \to \infty$ and then look for which $z$ this result will be $<1$. Those are the $z$ for which the series will converge (absolutely). – rehband Aug 21 '14 at 03:40
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    @rehband: gotcha. and it seems that besides applying the root test to the whole expression you can also do the limsup thing on the whole expression also. now I understand A) and B). your explanation is really appreciated. – mark leeds Aug 21 '14 at 04:10
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Regarding problem A:

Put $$a_n = \frac{n^n}{n!}$$ By the Cauchy D'Alembert criterion we have

$$\frac{1}{R}=\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{(n+1)^{n+1}}{(n+1)!}\frac{n!}{n^n}=\lim_{n\to\infty}\frac{(n+1)^n}{n^n}=\lim_{n\to\infty}\Big(1+\frac{1}{n}\Big)^n=e\\\iff R=\frac{1}{e}.$$

Regarding problem B:

Put $$b_n=z^{n!}.$$ Then $$\sqrt[n]{b_n}=|z|^{n!/n}$$ which converges to $\infty$ if $|z|>1$ and to $0$ if $|z|<1$. Thus the radius of convergence is $1$.

rehband
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