Answer:
∑ (-1)ⁿ⁺³ 1 / (n^½)
∑ (-1)³ⁿ 1 / (8 + n)
Explanation:
If ∑ an is convergent and ∑│an│is divergent, then the series is conditionally convergent.
Option A: (-1)²ⁿ is always +1. So an =│an│and both series converge (absolutely convergent).
Option B: bn = 1 / (n^⁹/₈) is a p series with p > 1, so both an and │an│converge (absolutely convergent).
Option C: an = 1 / n³ isn't an alternating series. So an =│an│and both series converge (p series with p > 1). This is absolutely convergent.
Option D: bn = 1 / (n^½) is a p series with p = ½, so this is a diverging series. Since lim(n→∞) bn = 0, and bn is decreasing, then an converges. So this is conditionally convergent.
Option E: (-1)³ⁿ = (-1)²ⁿ (-1)ⁿ = (-1)ⁿ, so this is an alternating series. bn = 1 / (8 + n), which diverges. Since lim(n→∞) bn = 0, and bn is decreasing, then an converges. So this is conditionally convergent.