Show that the following series converge on $\mathbb{C}$, $\sum_\limits{n=1}^{\infty}\frac{n!}{a^{n^2}}z^n$. $z\in\mathbb{C}$ is a variable while $a\in\mathbb{C}$ is a constant
I solved in the following way but a Professor told me I was wrong I did not understand why.
Using Weierstrass's test:
$\sum_\limits{n=1}^{\infty}\frac{n!}{a^{n^2}}z^n<\sum_\limits{n=1}^{\infty}\frac{n^n}{a^{n^2}}z^n$
Then I applied the root test to check if the series $\sum_\limits{n=1}^{\infty}\frac{n^n}{a^{n^2}}z^n$ converges:
$\lim_{n\to\infty}\sqrt[n]{\sum_\limits{n=1}^{\infty}\frac{n^n}{a^{n^2}}z^n}=\lim_{n\to\infty}\frac{1}{a^n ln(a)}z=0$ which proves that $\sum_\limits{n=1}^{\infty}\frac{n!}{a^{n^2}}z^n$ is convergent.
I was told to use an approximation of $n!$ to study the convergence of the series. I would find that the series would diverge.
Question:
Why is my method wrong? Am I applying the test in wrong way? Is Weierstrass´s test true on this case?
In the first place, you aren't using Weirstrass's M-test.$$\sum_\limits{n=1}^{\infty}\frac{n!}{a^{n^2}}|z|^n<\sum_\limits{n=1}^{\infty}\frac{n^n}{a^{n^2}}|z|^n$$ is certainly true, but it has nothing to do with the M-test. In the M-test you show that each summand $f_n$ is uniformly bound by some constant $M_n.$ You haven't done this, and in fact, it isn't true in this example.
Now when you are trying to use the root test, what you have is all wrong. You need to be taking the $n-$th root of the modulus of the $n-$th coefficient. What you have written, the square root of the sum make no sense, because we don't even know that the sum converges. If it does converge, it doesn't depend on $n,$ so taking the limit as $n\to\infty$ is pointless. What you want is $$\lim_{n\to\infty}\sqrt[n]{\frac{n^n}{a^{n^2}}}$$