Skip to main content

Recitation 20

Example 1: Suppose an\sum a_n and bn\sum b_n are series with positive terms and (a) If an>bna_n > b_n for all nn, what can you say about ana_n? Why? (b) If an<bna_n < b_n for all nn, what can you say about ana_n? Why?

Answer: (a) If bn\sum b_n diverges, so does an\sum a_n. (b) If bn\sum b_n converges, so does an\sum a_n.

Problem 2: Determine whether the series is convergent or divergent? (a) n=1n+1nn\sum_{n=1}^\infty \frac{n+1}{n\sqrt{n}}; (b) n=19n3+10n\sum_{n=1}^\infty\frac{9^n}{3+10^n}; (c) k=1ksin2k1+k3\sum_{k=1}^\infty\frac{k\sin^2k}{1+k^3}; (d) k=1(2k1)(k21)(k+1)(k2+4)2\sum_{k=1}^\infty\frac{(2k-1)(k^2-1)}{(k+1)(k^2+4)^2}; (e) n=14n+13n2\sum_{n=1}^\infty\frac{4^{n+1}}{3^n-2}; (f) n=11n2+1\sum_{n=1}^\infty\frac{1}{\sqrt{n^2+1}}; (g) n=11n!\sum_{n=1}^\infty\frac{1}{n!}; (h) n=1sin(1/n)\sum_{n=1}^\infty\sin(1/n); (i) n=11n1+1/n\sum_{n=1}^\infty\frac{1}{n^{1+1/n}}.

Hint: (a) Compare n+1nn\frac{n+1}{n\sqrt{n}} with 1/n1/\sqrt{n}; (b) Compare 9n3+10n\frac{9^n}{3+10^n} with (9/10)n(9/10)^n; (c) Use ksin2k1+k3<1/k2\frac{k\sin^2k}{1+k^3} < 1/k^2; (d) Compare (2k1)(k21)(k+1)(k2+4)2\frac{(2k-1)(k^2-1)}{(k+1)(k^2+4)^2} with 1/k31/k^3; (e) Compare 4n+13n2\frac{4^{n+1}}{3^n-2} with (4/3)n(4/3)^n; (f) Compare 1n2+1\frac{1}{\sqrt{n^2+1}} with 1/n1/n; (g) Use 1/n!<1/2n1/n! < 1/2^n for all n>1n>1; (h) Compare sin(1/n)\sin(1/n) with 1/n1/n; (i) Compare 1/n1+1/n1/n^{1+1/n} with 1/n1/n and use limnn1/n=1\lim_{n\to\infty}n^{1/n}=1.