Step-by-step explanation:
∑ sin(nπ/4) / (n³ + 3n)
This is less than 1 / n⁴ for all n > 1. 1/n⁴ is a convergent p-series, so the lesser series also converges. │aₙ│converges for the same reason, so this is absolutely convergent.
∑ (-1)ⁿ⁺¹ ln(n + 1) / (n + 1)
This is an alternating series. bₙ is positive and decreasing, and lim(n→∞) bₙ = 0, so the series converges. Now we need to check if│aₙ│converges. Using comparison test, │aₙ│is greater than 1/n for all n ≥ 6. 1/n is a divergent p-series, so the greater series│aₙ│also diverges. So this is conditionally convergent.