Vtyjkyuk

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
.

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 $-1 + sum_2^infty n^2 = +infty$.

Note that$$leftlvertfrac(n+1)^22^-(n+1)!x^(n+1)!n^22^-n!x^n!rightrvert=fracn+1nleftlvertfrac x2rightrvert^ntimes n!$$and that therefore$$lim_ntoinftyleftlvertfrac2^-(n+1)!x^(n+1)!2^-n!x^n!rightrvert=begincases0&text if |x|<2\+infty&text if |x|>2.endcases$$So, the radius of convergence is $2$.