Q:

What will be the interval of convergence $\sum_{n=1}^{\infty} \frac{n! x^n}{2^{n^{2}}}$

$\displaystyle \sum_{n=1}^{\infty} \frac{n! x^n}{2^{n^{2}}}$ My main problem is how can I handle the denominator ${2^{n^{2}}}$? I used until know the ratio test, as I saw a factorial, but it the denominator makes it a little bit difficult.

A:

With Hadamard's formula: the radius of convergence $R$ of $\sum a_n z^n$ is given by
$$\frac1R=\limsup\Bigl(\lvert a_n\rvert^{\tfrac1n}\Bigr).$$
$$\text{Now}\hspace{10em}0\le\biggl(\frac{n!}{2^{n^2}}\biggr)^{\tfrac1n}\le\biggl(\frac{n^n}{2^{n^2}}\biggr)^{\tfrac1n}=\frac n{2^n}\to 0\hspace{11em},$$
so $R=\infty$.

