radius of convergence of the series $\sum_{n=1}^{\infty} n^22^{\large -n!} x^{n!}$

Question

What is the radius of convergence of the series

$$ \sum_{n=1}^{\infty} n^22^{\large -n!} x^{n!} $$ ?

Answer:

If the series was

$$ \sum_{n=1}^{\infty} n^22^{\large -n!} x^{n} $$

then I could apply Hardamard Formula. But since there are $x^{n!}$ instead of $x^n$ I got stuck right here.

Help me doing this using Cauchy Hadamard formula .

Answer 1

Let $$ a_{n!} = n^2 \cdot 2^{-n!}, a_k = 0 \quad [k \notin \{j!\}_{j \in \mathbb N^*}], $$ then the series is $\sum a_j x^j$. Now use Cauchy-Hadamard as you want: $$ \frac 1R = \overline {\lim_k} |a_k|^{1/k} = \overline {\lim_n } |a_{n!}|^{1/n!} = \overline {\lim_n} (n^2)^{1/n!} 2^{-1} = \frac 12, $$ where $$ 1 \leqslant (n^2)^{1/n!} \leqslant (n^2)^{1/n^2} \to 1 \quad [n \to \infty]. $$ Hence $R = 2$.

UPDATE

At endpoints $\pm 2$, when $x = 2$, the series becomes $\sum n^2 = +\infty$. At $x = -2$, since $n!$ are even numbers for $n \geqslant 2$, the series becomes $-1 + \sum_2^\infty n^2 = +\infty$.

Answer 2

Note that$$\left\lvert\frac{(n+1)^22^{-(n+1)!}x^{(n+1)!}}{n^22^{-n!}x^{n!}}\right\rvert=\frac{n+1}n\left\lvert\frac x2\right\rvert^{n\times n!}$$and that therefore$$\lim_{n\to\infty}\left\lvert\frac{2^{-(n+1)!x^{(n+1)!}}}{2^{-n!}x^{n!}}\right\rvert=\begin{cases}0&\text{ if }|x|<2\\+\infty&\text{ if }|x|>2.\end{cases}$$So, the radius of convergence is $2$.