Question

How do you do this question?

Answer 1

Answer:

∑ (-1)ⁿ⁺³ 1 / (n^½)

∑ (-1)³ⁿ 1 / (8 + n)

Step-by-step 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.