How to calculate the radius and interval of convergence of the following series: $$\sum 2^{n^2}x^{n!}$$

The formula for the radius is: $$R = \frac{1}{\limsup_{n\to\infty} \sqrt[n]{|a_n|}}$$ or (that is if the limit exists): $$R = \lim_{n\to\infty}\bigg|\frac{a_n}{a_{n+1}}\bigg|$$

but as for now I always had series of the following form:

$$\sum a_n(x-c)^n$$

but now I have factorials $x^{n!}$ and $n^2$ in $a_n$, how do I approach this? The result itself is not very important to me but I want to understand how to approach problems where I have $x^{f(n)}$ instead of $x^n$.

Any hints, suggestions, etc. are welcome!


$$a_k = \begin{cases} 2^{n^2} &, k = n!\\ 0 &, \text{otherwise}\end{cases}$$

You have a lot of zero coefficients in the series. These don't matter for the radius of convergence, so you have

$$R = \frac{1}{\limsup \sqrt[n!]{2^{n^2}}} = 1.$$

If you have a series $$\sum_{n=0}^\infty c_n z^{a(n)}$$ with a strictly increasing $a\colon \mathbb{N}\to\mathbb{N}$, the radius of convergence is determined as

$$\frac{1}{R} = \limsup_{n\to\infty} \lvert c_n\rvert^{1/a(n)},$$

since only the nonzero coefficients matter.

  • $\begingroup$ ... and because $\lim_n a_n = \lim_n 2^{n^2} = + \infty \neq 0$ , the interval of convergence is $\langle -1, 1 \rangle$ and not $[-1, 1]$ ? $\endgroup$ – AltairAC Jun 15 '14 at 22:09
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    $\begingroup$ Yes. Since the terms of the series don't converge to $0$ for $\lvert x\rvert = 1$, the series converges only for $\lvert x\rvert < 1$. $\endgroup$ – Daniel Fischer Jun 15 '14 at 22:12
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    $\begingroup$ If the coefficients were $2^{-n^2}$ instead of $2^{n^2}$, the radius of convergence would also be $1$, but then the series would also converge for $\lvert x\rvert = 1$. $\endgroup$ – Daniel Fischer Jun 15 '14 at 22:14

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