Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the convergence or divergence of the series using other methods.\\
$\sum_{n=0}^{\infty} \frac{(n-1)!}{2^n}$\\Step 1\\Recall the Ratio Test states that if $\sum a_n$ is a series with nonzero terms and $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| < 1$, then $\sum a_n$ converges absolutely. If $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| > 1$ or $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \infty$, then $\sum a_n$ diverges\\Step 2\\For this series, $a_n = \frac{(n-1)!}{2^n}$.\\Find $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|$.\\$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{\frac{n!}{2^{n+1}}}{\frac{(n-1)!}{2^n}}| = \lim_{n \to \infty} |\frac{n!}{2^{n+1}} \cdot \frac{2^n}{(n-1)!}| = \lim_{n \to \infty} \frac{n}{2} = \infty$
