Answer:
Conditionally convergent
Step-by-step explanation:
Use alternating series test.
bₙ = n / √(n³ + 9)
lim(n→∞) bₙ
= lim(n→∞) [n / √(n³ + 9)]
= lim(n→∞) (n / √n³)
= lim(n→∞) (1 / √n)
= 0
The limit is 0, and bₙ is a decreasing series, so ∑aₙ converges.
Since bₙ > 1/n for n > 2, and 1/n diverges, ∑bₙ diverges.
(Alternatively, we could use limit comparison test to compare bₙ to n/√n³ and show that both diverge.)
Therefore, ∑aₙ is conditionally convergent.
