Question

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$

          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$

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.

∑n=0^∞((n-1)!)/(2^n)
Step 1
Recall the Ratio Test states that if ∑ an is a series with nonzero terms and limn →∞ |(an+1)/(an)| < 1, then ∑ an converges absolutely. If limn →∞ |(an+1)/(an)| > 1 or limn →∞ |(an+1)/(an)| = ∞, then ∑ an diverges
Step 2
For this series, an = ((n-1)!)/(2^n).
Find limn →∞ |(an+1)/(an)|.
limn →∞ |(an+1)/(an)| = limn →∞ |((n!)/(2^n+1))/(((n-1)!)/(2^n))| = limn →∞ |(n!)/(2^n+1)·(2^n)/((n-1)!)| = limn →∞(n)/(2) = ∞

Added by Ryan S.

Question

Please give Ace some feedback