Use the Ratio Test to determine whether the series is convergent or divergent.

Σ[infinity] n = 1 (-1)^n - 1 3^n/2^nn^3

The series is absolutely convergent.

Step-by-step explanation:

By ratio test, we find the limit as n approaches infinity of

|[a_(n+1)]/a_n|

a_n = (-1)^(n - 1).(3^n)/(2^n.n^3)

a_(n+1) = (-1)^n.3^(n+1)/(2^(n+1).(n+1)^3)

[a_(n+1)]/a_n = [(-1)^n.3^(n+1)/(2^(n+1).(n+1)^3)] × [(2^n.n^3)/(-1)^(n - 1).(3^n)]

= |-3n³/2(n+1)³|

= 3n³/2(n+1)³

= (3/2)[1/(1 + 1/n)³]

Now, we take the limit of (3/2)[1/(1 + 1/n)³] as n approaches infinity

= (3/2)limit of [1/(1 + 1/n)³] as n approaches infinity

= 3/2 × 1

= 3/2

The series is therefore, absolutely convergent, and the limit is 3/2

1/4 because there is 4 grades.