Example 1: Suppose \sum a_n and \sum b_n are series with positive terms and (a) If a_n > b_n for all n, what can you say about a_n? Why? (b) If a_n < b_n for all n, what can you say about a_n? Why?
Answer: (a) If \sum b_n diverges, so does \sum a_n. (b) If \sum b_n converges, so does \sum a_n.
Problem 2: Determine whether the series is convergent or divergent? (a) \sum_{n=1}^\infty \frac{n+1}{n\sqrt{n}}; (b) \sum_{n=1}^\infty\frac{9^n}{3+10^n}; (c) \sum_{k=1}^\infty\frac{k\sin^2k}{1+k^3}; (d) \sum_{k=1}^\infty\frac{(2k-1)(k^2-1)}{(k+1)(k^2+4)^2}; (e) \sum_{n=1}^\infty\frac{4^{n+1}}{3^n-2}; (f) \sum_{n=1}^\infty\frac{1}{\sqrt{n^2+1}}; (g) \sum_{n=1}^\infty\frac{1}{n!}; (h) \sum_{n=1}^\infty\sin(1/n); (i) \sum_{n=1}^\infty\frac{1}{n^{1+1/n}}.
Hint: (a) Compare \frac{n+1}{n\sqrt{n}} with 1/\sqrt{n}; (b) Compare \frac{9^n}{3+10^n} with (9/10)^n; (c) Use \frac{k\sin^2k}{1+k^3} < 1/k^2; (d) Compare \frac{(2k-1)(k^2-1)}{(k+1)(k^2+4)^2} with 1/k^3; (e) Compare \frac{4^{n+1}}{3^n-2} with (4/3)^n; (f) Compare \frac{1}{\sqrt{n^2+1}} with 1/n; (g) Use 1/n! < 1/2^n for all n>1; (h) Compare \sin(1/n) with 1/n; (i) Compare 1/n^{1+1/n} with 1/n and use \lim_{n\to\infty}n^{1/n}=1.