Radius of convergence of the series $\displaystyle\sum_{n=0}^{\infty}n!z^{2n+1}$ 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
 A: 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.
A: 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 $R^2$ .
A: By Cauchy-Hadamard,  $r=\dfrac1{\limsup_{n\to\infty}\sqrt[2n+1]{n!}}=0$.
Use Stirling's approximation .  
