Answer: To apply the Ratio Test to the series
∞Σn=1 (-1)^n 2^(n) n / (5 · 8 · 11 · ... · (3n - 2))
we need to compute the limit of the ratio of successive terms:
|a_{n+1}| / |an| = [(2^(n+1))(n+1)] / [(3n+1)(3n+2)(3n+3)]
Simplifying this expression, we get:
|a_{n+1}| / |an| = [(2n+2)/3] / [(3n+1)(3n+2)/3]
|a_{n+1}| / |an| = (2n+2)/(9n^2 + 11n + 2)
Now, taking the limit as n approaches infinity:
lim n → ∞ |a_{n+1}| / |an| = lim n → ∞ (2n+2)/(9n^2 + 11n + 2)
Since the degree of the numerator and denominator are equal, we can apply L'Hopital's rule:
lim n → ∞ |a_{n+1}| / |an| = lim n → ∞ (2/(18n+11)) = 0
Since the limit of the ratio is less than 1, by the Ratio Test, the series is absolutely convergent. Therefore, the series converges.