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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?

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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 $\vert z\vert\gt1$.

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  • $\begingroup$ When $|z|>1$ then my limit becomes $0.\infty=0$ in $\Bbb C_{\infty}$. $\endgroup$ – Empty Jul 20 at 19:13
  • $\begingroup$ How your limit becomes $1$? I'm not getting. Can you please give some detail? $\endgroup$ – Empty Jul 20 at 19:14
  • $\begingroup$ It took awhile, and i used an online calculator. $\endgroup$ – Chris Custer Jul 20 at 19:23
  • $\begingroup$ Exponentials grow faster than polynomials. That's why I think your ratio test gives $\infty$. $\endgroup$ – Chris Custer Jul 20 at 19:43
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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 $\:o\bigl(|z|^{3n(n+1)+1}\bigr)$.

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