Radius of convergence of the series $\displaystyle\sum_{n=0}^{\infty}n!z^{2n+1}$

Question

We have the result, $\displaystyle{\frac{1}{R}=lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|}$, where $R$ is the radius of the convergence, where $\displaystyle{a_n}$ is the coefficient of the series $\displaystyle{\sum_{n=0}^{\infty}}a_nz^n$,

Here we redefine the series as$\displaystyle{\sum_{n=0}^{\infty}}a_nz^n$, where $a_n=(0,1,0,1,0,2,0,6,0,24,0,.....)$, so we can not use this method? or can we?,

We have other result as $\displaystyle{\frac{1}{R}=\lim \inf} |a_n|^{-1/n}=\lim\inf{\frac{1}{|a_n|^{1/n}}}$

My question is

1)Is it valid to take $a_n=0$ for infinitely many $n\in \mathbb{N}$

2) $\displaystyle{\frac{1}{R}=\lim\inf{\frac{1}{|n!|^{1/n}}}}????$

Can someone help how to move further

Answer 1

First a comment

You made an error in your question. You have $$\displaystyle{\frac{1}{R}=\limsup} |a_n|^{-1/n}$$ and not

$$\displaystyle{\frac{1}{R}=\liminf} |a_n|^{-1/n}$$

Then I would proceed like that for the radius of convergence.

I would say $$f(z)=\displaystyle\sum_{n=0}^{\infty}n!z^{2n+1} = z \displaystyle\sum_{n=0}^{\infty}n!(z^2)^n$$

So if $u_n(z) = n!(z^2)^n$, you have

$$\left\vert \frac{u_{n+1}(z)}{u_n(z)} \right\vert = n\vert z \vert^2$$

So apart from $z = 0$, $\lim\limits_{n \to \infty} \left\vert \frac{u_{n+1}(z)}{u_n(z)} \right\vert = \infty$. As the term of the series $\sum u_n(z)$ doesn't converge to $0$, the series diverges. Hence the radius of convergence of $f$ is equal to zero.

Answer 2

You can use

$$S(z)=\sum_{n=0}^\infty n!z^{2n+1}=z\sum_{n=0}^\infty n!(z^2)^n$$ and study the convergence of the series $$T(w)=\sum_{n=0}^\infty n!w^n.$$

If the radius of convergence of $S$ (for $z$) is $R$, then that of $T$ (for $w$) is $R^2$ .

Answer 3

By Cauchy-Hadamard, $r=\dfrac1{\limsup_{n\to\infty}\sqrt[2n+1]{n!}}=0$.

Use Stirling's approximation .