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

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

## marked as duplicate by Jyrki Lahtonen, Community♦Sep 6 at 16:46

• The "result" only holds, when the limit exists, because in general, it's $\lim \sup$ and not $\lim$. – Botond Sep 6 at 16:20
• What about dropping zeros and writing the series as $$\sum_{n=0}^{\infty}a_{2n}z^{2n}$$? – Mostafa Ayaz Sep 6 at 16:21
• Many related questions. You really should search the site before asking. The same applies to the answerers. – Jyrki Lahtonen Sep 6 at 16:29

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 $$R^2$$ .

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.

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