1
$\begingroup$

Let's suppose $R_1>0$ radius of convergence of the power serie $\sum_{n=0}^\infty a_nz^n$. What is the convergence radius of the series $\sum_{n=0}^\infty\frac{a_n}{n!}z^n$?

Idea: By Cauchy-Hadamard theorem $\frac{1}{R_2}=\limsup_{n\rightarrow \infty} \sqrt[n]{|b_n|}$ with $R_2$ radius of convergence of the power serie $\sum_{n=0}^\infty\frac{a_n}{n!}z^n$ and $b_n=\frac{a_n}{n!}$. Then...

$\frac{1}{R_2}=\limsup_{n\rightarrow \infty} \sqrt[n]{|b_n|}=\limsup_{n\rightarrow \infty} \sqrt[n]{|\frac{a_n}{n!}|}=\limsup_{n\rightarrow \infty} \frac{\sqrt[n]{|a_n|}}{\sqrt[n]{n!}}=\frac{\limsup_{n\rightarrow \infty}\sqrt[n]{|a_n|}}{\limsup_{n\rightarrow \infty}\sqrt[n]{n!}}???$

can I assure that ${\{|a_n|}\}_{n\in\mathbb{N}}$ converges?

$R_2=\infty$?

Note: To apply the quotient critic $\lim_{n\rightarrow \infty}|\frac{b_n}{b_{n+1}}|$ we need to ${\{n\in\mathbb{N}:b_n=0}\} $ finite

Can someone help me solve the problem?

$\endgroup$
3
  • 4
    $\begingroup$ It looks fine to me: if $\sum a_n z^n$ has a positive radius of convergence, $\sum\frac{a_n}{n!}z^n$ is an entire function. $\endgroup$ Sep 4, 2019 at 20:56
  • 2
    $\begingroup$ I’m not sure you’ve specified the details that let you say “hence...” but it is true that $R_2=\infty.$ $\endgroup$ Sep 4, 2019 at 21:00
  • $\begingroup$ @JackD'Aurizio Yes. But, How do I do it with the criteria? $\endgroup$ Sep 4, 2019 at 21:33

1 Answer 1

2
$\begingroup$

1) Your last equality is not justified: it is not always true that $\limsup \dfrac{a_n}{b_n}=\dfrac{\limsup a_n}{\limsup b_n}$. Instead, you'll have to argue that the numerator of your fraction is bounded (since it has a finite $\limsup$) and the denominator diverges to $+\infty$.

Mouseover for hint #2:

2) To show that $\sqrt[n]{n!}$ diverges to $+\infty$, begin by noting that $n!>\left\lfloor\frac{n}{2}\right\rfloor^{n/2}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .