Question

Classify And Test Each Series For Convergence Or Divergence

a) ∑N=1[infinity]1+2n∣Sin2n∣
b) ∑N=1[infinity]Sin(1/N)

Answer 1

a) The series ∑N=1∞(1+2n)|Sin2n| diverges. b) The series ∑N=1∞Sin(1/N) converges.

Step-by-step explanation:

a) The series ∑N=1∞(1+2n)|Sin2n| can be split into two parts: the first part is the series ∑N=1∞(1+2n), and the second part is the series ∑N=1∞|Sin2n|. The first part diverges because the sum of (1+2n) grows without bound as n increases. The second part also diverges because the absolute value of Sin2n does not approach zero as n goes to infinity. Therefore, the overall series also diverges.

b) The series ∑N=1∞Sin(1/N) converges. The series is a converging alternating series, where each term approaches zero as n approaches infinity. Therefore, the series converges by the Alternating Series Test.