Find the radius of convergence of the power series $\sum_{n=1}^\infty 2^{n^2} \: x^{n!}$

Question

This is a problem from the book of Kaczor & Nowak (Version II).
In the hint part, they find the radius of convergence by calculating the limsup of $(2^{n^2})^{1 \over n!}$. But I am unable to understand why not limsup of $(2^{n^2})^{1 \over n}$?
Can anybody please tell me how to find the radius of convergence of this kind of power series where the power of $x$ is not $n$ (or power of $x$ is an injective function of $n$) with a rigorous analytical argument?

Answer 1

Just write the power series in the form $\sum_{n=0}^\infty a_nx^n$. In this case we have

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

so

$$\frac1{R} = \limsup_{n\to\infty} {a_n}^{\frac1n} = \limsup_{k\to\infty} {a_{k!}}^{\frac1{k!}} = \limsup_{k\to\infty} {\left(2^{k^2}\right)}^{\frac1{k!}} = \limsup_{k\to\infty} 2^{\frac{k^2}{k!}} = 1$$

because we took the limsup of the only nontrivial subsequence of ${a_n}^{\frac1n}$, which is ${a_{k!}}^{\frac1{k!}}$ .

Answer 2

Well, $x^{n!} = 2^{n!\log(x)}$ which for $x<1$ is $2^{-an!}$ for some positive $a(x)=a$ that is independent of $n$.

Note that for any such $a$, the function $an!$ dominates $n^2$ for $n$ large.

MEanwhile 0 raised to any positive power is 0.

Furthermore, one can check that if the sum converges for $x \in [0,1)$ then it also converges for $-x$.

On the other hand, $n!$ is even for all $n \geq 2$. So clearly this diverges for $x=-1$. (And clearly this diverges for $x \ge 1$.)

So the sum converges for all $x \in (-1,1)$. Thus the radius of convergence is 1.

Answer 3

Note that for $|x|\ge 1$

$$|2^{n^2} \: x^{n!}|\to\infty$$

for $|x|< 1$ let $y=\frac 1 x$

$$\left|\frac{2^{n^2}}{y^{n!}}\right|\to0$$

and $\sum \frac{2^{n^2}}{y^{n!}}$ converges by limit comparison test with $\sum \frac 1 {n^2}$.

Then the radius of convergence is $R=1$.