From the James Stewart Essential Calculus Early Transcendentals Second Edition textbook I have the given problem:
$$a_n=\frac{(-3)^n}{n!}$$
I did the following absolute convergence procedure, because I treated the problem as the following:
Theorem 6
If $ \lim_\limits{n \to \infty}\rvert a_n \lvert = 0$, $\ \therefore \ $ $\lim_\limits{n \to \infty} a_n = 0$ $$\lim_\limits{n \to \infty} \frac{(-1)^n (3)^n}{n!}$$ I then then proceeded to do the following, in accordance to the theorem, and the way which I manipulated the sequence. \begin{align} \lim_\limits{n \to \infty} \vert \frac{(-1)^n (3)^n}{n!} \vert \\ \lim_\limits{n \to \infty} \frac{ (3)^n}{n!} \end{align}
Here is where I get a bit lost because of the intuition behind the problem, I graphed this as function, and determined the limit equals 0, but I am trying to determine it analytically. I went to Symbolab, and it says that numerator is undefined, and that its denominator is infinity, thus it being 0. But this is how I view the problem.
My Work
$$\lim_\limits{n \to \infty} \frac{ (3)^n}{n!}$$ $$\frac{3^\infty}{\infty!}$$ My thoughts were that the $3^\infty = \infty$, thus being $$\frac{\infty}{\infty}$$. Is there something I am missing, because L^Hopital's rule is not applicable here. How do I analytically get at the problem?
