# Find Radius of convergence of the given power series

Find the radius of convergence of the power series $$\displaystyle \sum_{n=1}^{\infty} \frac{n+1}{n!} z^{n^3}$$.

Let, $$\displaystyle a_n=\frac{n+1}{n!} z^{n^3}$$.

Then, $$\displaystyle \left|\frac{a_{n+1}}{a_n}\right|=\frac{n+2}{(n+1)^2}|z|^{3n(n+1)+1}\longrightarrow 0$$ for all $$z\in \Bbb C$$. So radius of convergence is $$\infty$$. Is it correct?

Using Cauchy-Hadamard, I get $$r=\dfrac 1{\limsup_{n\to\infty}\sqrt[n^3]{\dfrac{n+1}{n!}}}=1$$.
I think your answer above diverges when $$\vert z\vert\gt1$$.
• When $|z|>1$ then my limit becomes $0.\infty=0$ in $\Bbb C_{\infty}$. – Empty Jul 20 at 19:13
• How your limit becomes $1$? I'm not getting. Can you please give some detail? – Empty Jul 20 at 19:14
• Exponentials grow faster than polynomials. That's why I think your ratio test gives $\infty$. – Chris Custer Jul 20 at 19:43
Your limit with the ratio test, tends to $$0$$ only if $$|z|\le 1$$. Indeed, if $$|z|>1$$, any polynomial $$p(n)$$ is $$o\bigl(|z|^n\bigr)$$, and a fortiori is $$\:o\bigl(|z|^{3n(n+1)+1}\bigr)$$.