Determine the number of terms sufficient to obtain the sum of the series accurate to three decimal places. $\sum _ { n = 0 } ^ { \infty } \frac { ( - 5 ) ^ { n + 3 } } { ( n + 1 ) ! }$
A) 19
B) 22
C) 20
D) 21

Question

Quizplus · Accepted Answer

To determine the number of terms needed for the series to converge to three decimal places, we need to find the smallest $ n $ such that the absolute value of the next term is less than 0.0005. The series is $\sum _ { n = 0 } ^ { \infty } \frac { ( - 5 ) ^ { n + 3 } } { ( n + 1 ) ! }$, which simplifies to $\sum _ { n = 0 } ^ { \infty } \frac { ( - 5 ) ^ { n } } { ( n + 4 ) ! }$. The term $\frac { ( - 5 ) ^ { n } } { ( n + 4 ) ! }$ should be less than 0.0005. Calculating for different $ n $, we find that $ n = 20 $ is the smallest value satisfying this condition.

Determine the Number of Terms Sufficient to Obtain the Sum ∑n=0∞(−5)n+3(n+1)!\sum _ { n = 0 } ^ { \infty } \frac { ( - 5 ) ^ { n + 3 } } { ( n + 1 ) ! }∑n=0∞​(n+1)!(−5)n+3​

Determine the Number of Terms Sufficient to Obtain the Sum $\sum _ { n = 0 } ^ { \infty } \frac { ( - 5 ) ^ { n + 3 } } { ( n + 1 ) ! }$