1. The series converges absolutely.

1. To test absolute convergence, we test the series: 1 + |-1/3| + 1/9 + |-1/27| + 1/81 + … the geometric series with r = 1/3. This series converges, so the alternating series converges absolutely.

1. The series converges conditionally.

1. To test absolute convergence, test the series with terms |a_n| = n/(n² + 2). The terms look like n/n2 = 1/n, which diverges. Because n/(n² + 2) < 1/n, we cannot apply the Comparison Test. Instead, apply the Limit Comparison Test. lim_{n → ∞} n/(n² + 2)/(1/n)= lim_{n → ∞} 1/(1 + 2/n²) = 1 Consequently, the series with terms |an| diverges. To test conditional convergence, we apply the Alternating Series Test. Consider the function f(x)= x/(x² + 2) and compute f'(x)= (2 – x²)/(x² + 2)². Because f'(x) is negative for x > √2, the terms |a_n| are eventually decreasing. Also, lim_{x → ∞} f(x)= 0, so lim_{n → ∞,/sub> an= 0. Then by the Alternating Series Test, the series converges. Because it does not converge absolutely, the series converges conditionally.}

1. The series diverges.

1. The numerator and denominator have the same highest power of n, so apply the nth term test lim_{n → ∞}(n² – 2n + 3)/(n² + 2n + 5) = lim_{n → ∞}(1 – 2/n + 3/n²)/(1 + 2/n + 5/n²)= 1 Because the terms do not go to 0, the series diverges.

1. The series converges absolutely.

1. Note |a_n| ≤ 1/n², a p-series with p= 2, converging by the Integral Test. Because the series with term |a_n| converges, the original series converges absolutely.

1. Converges absolutely for p > 1, converges conditionally for 0 < p ≤ 1, and diverges for p ≤ 0.

1. To test absolute convergence, we test the series: 1 + |-1/2^p| + |1/3^p| + |-1/4^p| + … the regular p-series. From the Integral Test, we know the p-series converges for p > 1 and diverges for p ≤ 1. Consequently, the alternating p-series converges absolutely for p > 1. For 0 < p ≤ 1, apply the Alternating Series Test. For f(x)= 1/x^p, we find f'(x)= -p/x^p+1 so f(x) is decreasing. Also, lim_{n → ∞} 1/n^p= 0 so the alternating p-series converges. Because the series does not converge absolutely in this range of p-values, the series converges conditionally. For p ≤ 0, the series diverges by the n^th term test.