Use the Ratio Test to determine if the series converges or diverges.
-$$\sum _ { n = 1 } ^ { \infty } \frac { 4 n ! } { n ^ { n } }$$
A) Diverges
B) Converges

Question

Quizplus · Accepted Answer

Let $a_n = \frac{4n!}{n^n}$.
Using the Ratio Test, we have
\begin{align*}
\lim_{n	o\infty} \left| \frac{a_{n+1}}{a_n} ight| &= \lim_{n	o\infty} \left| \frac{4(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{4n!} ight| \
&= \lim_{n	o\infty} \left| \frac{4(n+1)n^n}{(n+1)^{n+1}} ight| \
&= \lim_{n	o\infty} \frac{4n^n}{(n+1)^n} \
&= \lim_{n	o\infty} \frac{4}{(1+\frac{1}{n})^n} \
&= 4e^{-1} \
&< 1.
\end{align*}
Therefore, the series $\sum_{n=1}^\infty \frac{4n!}{n^n}$ converges by the Ratio Test. The best choice is $\boxed{	ext{(B) Converges}}$.

Use the Ratio Test to Determine If the Series Converges ∑n=1∞4n!nn\sum _ { n = 1 } ^ { \infty } \frac { 4 n ! } { n ^ { n } }n=1∑∞​nn4n!​

Use the Ratio Test to Determine If the Series Converges $\sum _ { n = 1 } ^ { \infty } \frac { 4 n ! } { n ^ { n } }$