The comparison test (see page 731) states that if and are series with positive terms then (i) If is conv

Question

The comparison test (see page 731) states that if and are series with positive terms then (i) If is convergent and for all , then is also convergent. (ii) If is divergent and for all , then is also divergent. Use the comparison test to determine the convergence or divergence of the following series:

(a) ∑n=1 ∞ (1/√n) b)∑n=1 ∞ 1/(n +2n) (c)∑n=1 ∞ n/(2n3 + 1)

Answer 1

Some of the stuff didn't come through on my machine, but:
(a) For all n ∈ Z+, 1/√n >= 1/n. Since Σ(n=1 to ∞) 1/n diverges, Σ(n=1 to ∞) 1/√n also diverges.

(b) I'm not sure that you've copied this question correctly - or perhaps you used some characters which didn't come across. It reads to me as 1/(n+2n), i.e. 1/(3n) which clearly diverges since it's just a constant multiple of 1/n. Perhaps it should be 1/(n^2 + 2n)? In this case, 1/(n^2+2n) < 1/n^2 and Σ(n=1 to ∞) converges, so Σ(n=1 to ∞) 1/(n^2 + 2n) converges.

(c) n / (2n^3 + 1) < n / (2n^3) = 1 / (2n^2) < 1 / n^2. Since Σ(n=1 to ∞) 1/n^2 converges, so does Σ(n=1 to ∞) n / (2n^3 + 1).