Use the Ratio Test to determine if the series converges or diverges.
-$$\sum _ { n = 1 } ^ { \infty } \frac { 8 ^ { n } } { n ! }$$
A) Diverges
B) Converges

Question

Quizplus · Accepted Answer

The Ratio Test states that if $\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=L$, then the series converges if $0 \leq L < 1$ and diverges if $L > 1$.

In this case, $a_n=\frac{8^n}{n!}$, so $\frac{a_{n+1}}{a_n}=\frac{8^{n+1}}{(n+1)!}\cdot\frac{n!}{8^n}=\frac{8}{n+1}$. Taking the limit as $n\rightarrow\infty$, we get $\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n\rightarrow\infty}\frac{8}{n+1}=0$. Since $0 \leq 0 < 1$, the series converges by the Ratio Test.

Use the Ratio Test to Determine If the Series Converges ∑n=1∞8nn!\sum _ { n = 1 } ^ { \infty } \frac { 8 ^ { n } } { n ! }n=1∑∞​n!8n​

Use the Ratio Test to Determine If the Series Converges $\sum _ { n = 1 } ^ { \infty } \frac { 8 ^ { n } } { n ! }$