# What is the convergence radius of the series $\sum_{n=0}^\infty\frac{a_n}{n!}z^n$?

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?

• 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. Sep 4, 2019 at 20:56
• I’m not sure you’ve specified the details that let you say “hence...” but it is true that $R_2=\infty.$ Sep 4, 2019 at 21:00
• @JackD'Aurizio Yes. But, How do I do it with the criteria? Sep 4, 2019 at 21:33

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