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