Part I Multiple Choice
(A) If both ∑n=1∞an and ∑n=1∞bn are absolutely convergent, then ∑n=1∞anbn=(∑n=1∞an)⋅(∑n=1∞bn).
(B) If ∑n=1∞an is convergent and ∑n=1∞bn is divergent, then ∑n=1∞(an+bn) must be divergent.
(C) Suppose that f(x) is a positive and continuous function on [1,∞), and the improper integral ∫1∞f(x)dx is convergent. Let an=f(n), then ∑n=1∞an is convergent.
(D) If an≤bn for all n∈N and ∑n=1∞bn is convergent, then ∑n=1∞an must be convergent.