Use the Ratio Test to determine the convergence or divergence of the series $$\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } ( 7 ) ^ { 8 n } } { ( 7 n + 1 ) ! }$$
A) diverges
B) converges

Question

Quizplus · Accepted Answer

We apply the ratio test: 
\begin{align*}
\lim_{n	o\infty}\frac{\left\vert \frac{(-1)^{n+1}(7)^{8(n+1)}}{(7(n+1)+1)!}ight\vert}{\left\vert \frac{(-1)^{n}(7)^{8n}}{(7n+1)!}ight\vert}
&=\lim_{n	o\infty}\frac{(7)^8}{(7n+8)(7n+1)}\
&=\lim_{n	o\infty}\frac{(7)^8/n^2}{49+8/n}\
&=0<1
\end{align*}
Hence, the series converges by the ratio test.

Use the Ratio Test to Determine the Convergence or Divergence ∑n=0∞(−1)n(7)8n(7n+1)!\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } ( 7 ) ^ { 8 n } } { ( 7 n + 1 ) ! }n=0∑∞​(7n+1)!(−1)n(7)8n​

Use the Ratio Test to Determine the Convergence or Divergence $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } ( 7 ) ^ { 8 n } } { ( 7 n + 1 ) ! }$