00 2 ln k k (2) Σ k=1 O 2 In k k 2 ln k k 2 In k k 2 In k k 1 k > 00 k > 2, Σ = diverges, Σ k k=1 k=1 1 k for k ≥ 2, Σ k=1 1 1 <= for k ≥ 2, Σ converges, k k k=1 1 k for for k ≥ 2, converges, Σ k=1 1 Σε k k=1 2 In k k k=1 diverges, Σ k=1 diverges 2 In k k 2 In k k 2 ln k k converges converges diverges

1 for convergence, you 2k5 +9k8 will use the Limit Comparison Test, comparing it to the p 1 series 9k² To test the series k=1 k=1 k=1 where p Now by the limit comparison test, the series ∞ 1 2k5 +9k8 = diverges converges O

7" (f) Ž F(-8)*| n-1 n=1 Converges or Diverges: Sum(If Converges): [1-(-1)" | (g) n+1 Converges or Diverges: limit (If Converges): (h) ¿l-(-1)" n+1 n=1 Converges or Diverges: Sum(If Converges): 5 (i) -2, 2' 25 125 8 32 Converges or Diverges: limit (If Converges): 5 (j) -2+ 25 125 8 32 Converges or Diverges: Sum(If Converges):

not graded question, do not reject