Test the series $\sum _ { n = 1 } ^ { \infty } \frac { n 4 ^ { n } } { n ! }$ for convergence or divergence using any appropriate test.
A)diverges
B)converges

Question

Quizplus · Accepted Answer

The series converges by the Ratio Test. For the given series, let $a_n = \frac{n4^n}{n!}$. The Ratio Test involves calculating $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$. Here, $\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1)4^{n+1}}{(n+1)!} \cdot \frac{n!}{n4^n} \right| = \left| \frac{4}{1} \right| = 4 \cdot \frac{1}{n+1}$, which approaches 0 as $n$ approaches infinity. Since this limit is less than 1, the series converges by the Ratio Test.

Test the Series ∑n=1∞n4nn!\sum _ { n = 1 } ^ { \infty } \frac { n 4 ^ { n } } { n ! }∑n=1∞​n!n4n​

Test the Series $\sum _ { n = 1 } ^ { \infty } \frac { n 4 ^ { n } } { n ! }$