Is the radius of convergence of the Taylor series of $f(x)=e^{x^4}$centered at $x_0=0$

Question

Am I right that the radius of convergence of the Taylor series of $f(x)=e^{x^4}$centered at $x_0=0$ is infinity since there are no singularities?

note: x is real valued.

Thanks!

Answer 1

$$ e^{x^4}=\sum_{k\geq 0}\frac{x^{4k}}{k!} $$ and you require $$ x^{4}\lim_{k\to \infty}\sup\left | \frac{1}{(k!)^{1/k}} \right|<1 $$ but that limit supremum is zero, and you may conclude that any $x$ will do.

Also note that this should be somewhat clear a priori; the factorial beats polynomial growth quite handily.

Answer 2

Note that the radius of convergence of $$e^t = \sum _{0}^{\infty} \frac {t^n}{n!} $$is $\infty$.

Thus the series will converge at $t=x^4$ as well.

That is the radius of convergence of $$e^{x^4} = \sum _{0}^{\infty} \frac {x^{4n}}{n!}$$ is $\infty$.