Infi-1 9

1a

\begin{matrix} Let \sum_{n = 1}^{\infty} a_{n}, \sum_{n = 1}^{\infty} b_{n} > 0 diverge \\ Prove or disprove: \sum_{n = 1}^{\infty} a_{n} b_{n} diverges \\ Disproof: \\ Let a_{n} = b_{n} = \frac{1}{n} \\ \sum_{n = 1}^{\infty} \frac{1}{n} diverges \\ \sum_{n = 1}^{\infty} a_{n} b_{n} = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} converges \end{matrix}

1b

\begin{matrix} Let \sum_{n = 1}^{\infty} a_{n}, \sum_{n = 1}^{\infty} b_{n} > 0 \\ \sum_{n = 1}^{\infty} a_{n}^{2}, \sum_{n = 1}^{\infty} b_{n}^{2} converge \\ Prove or disprove: \sum_{n = 1}^{\infty} a_{n} b_{n} converges \\ Proof: \\ (a - b)^{2} \geq 0 ⟹ a^{2} + b^{2} \geq 2 a b ⟹ \frac{a^{2} + b^{2}}{2} \geq a b \\ Let c_{n} = \frac{a_{n}^{2} + b_{n}^{2}}{2} \\ \sum_{n = 1}^{\infty} a_{n}^{2} converges \\ \sum_{n = 1}^{\infty} b_{n}^{2} converges \\ ⟹ \sum_{n = 1}^{\infty} c_{n} = \frac{1}{2} \sum_{n = 1}^{\infty} (a_{n}^{2} + b_{n}^{2}) converges \\ c_{n} \geq a_{n} b_{n} and \sum_{n = 1}^{\infty} c_{n} converges \\ ⟹ By the direct comparison test: \sum_{n = 1}^{\infty} a_{n} b_{n} converges \end{matrix}

1c

\begin{matrix} Let \sum_{n = 1}^{\infty} a_{n}^{2} converge \\ Prove or disprove: \sum_{n = 1}^{\infty} \frac{a_{n}}{n} converges absolutely \\ Proof: \\ Let b_{n} = \frac{1}{n} \\ \sum_{n = 1}^{\infty} | a_{n} | > 0 \\ \sum_{n = 1}^{\infty} (| a_{n} |)^{2} = \sum_{n = 1}^{\infty} a_{n}^{2} converges \\ \sum_{n = 1}^{\infty} b_{n}^{2} = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} converges \\ ⟹ \sum_{n = 1}^{\infty} | \frac{a_{n}}{n} | = \sum_{n = 1}^{\infty} | a_{n} | \frac{1}{n} = \sum_{n = 1}^{\infty} | a_{n} | b_{n} converges as proved in 1b \\ ⟹ \sum_{n = 1}^{\infty} \frac{a_{n}}{n} converges absolutely \end{matrix}

2

\begin{matrix} Determine whether series converges absolutely, conditionally or diverges \end{matrix}

2a

\begin{matrix} \sum_{n = 1}^{\infty} \frac{\sin (n) \cos (n) \sin (n^{3})}{n \sqrt{n}} \\ Solution: \\ | \sin (n) \cos (n) \sin (n^{3}) | \leq 1 \\ ⟹ | \frac{\sin (n) \cos (n) \sin (n^{3})}{n \sqrt{n}} | \leq \frac{1}{n \sqrt{n}} \\ \sum_{n = 1}^{\infty} \frac{1}{n \sqrt{n}} converges \\ ⟹ By the direct comparison test: \sum_{n = 1}^{\infty} | \frac{\sin (n) \cos (n) \sin (n^{3})}{n \sqrt{n}} | converges \\ ⟹ \sum_{n = 1}^{\infty} \frac{\sin (n) \cos (n) \sin (n^{3})}{n \sqrt{n}} converges absolutely \end{matrix}

2b

\begin{matrix} \sum_{n = 1}^{\infty} (- 1)^{n + 1} \frac{n!}{3^{n}} \\ Solution: \\ lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = lim_{n \to \infty} \frac{(n + 1)! \cdot 3^{n}}{n! \cdot 3^{n + 1}} = lim_{n \to \infty} \frac{n + 1}{3} = \infty > 1 \\ ⟹ By the ratio test: \sum_{n = 1}^{\infty} (- 1)^{n + 1} \frac{n!}{3^{n}} diverges \end{matrix}

2c

\begin{matrix} \sum_{n = 1}^{\infty} (- 1)^{n} \frac{(n!)^{3}}{(3 n)!} \\ Solution: \\ lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = lim_{n \to \infty} \frac{((n + 1)!)^{3} \cdot (3 n)!}{(3 n + 3)! \cdot (n!)^{3}} = lim_{n \to \infty} \frac{(n + 1)^{3}}{(3 n + 3) (3 n + 2) (3 n + 1)} = \\ = lim_{n \to \infty} \frac{(n + 1)^{2}}{3 (3 n + 2) (3 n + 1)} = lim_{n \to \infty} \frac{n^{2} + 2 n + 1}{27 n^{2} + 27 n + 9} = \frac{1}{27} < 1 \\ ⟹ By the ratio test: \sum_{n = 1}^{\infty} (- 1)^{n} \frac{(n!)^{3}}{(3 n)!} converges absolutely \end{matrix}

2d

\begin{matrix} \sum_{n = 1}^{\infty} \frac{(- 1)^{n}}{\ln (n!)} \\ Solution: \\ \sum_{n = 1}^{\infty} | \frac{(- 1)^{n}}{\ln (n!)} | = \sum_{n = 1}^{\infty} \frac{1}{\ln (n!)} \\ n^{n} > n! ⟹ \ln (n^{n}) > \ln (n!) ⟹ \frac{1}{\ln (n^{n})} < \frac{1}{\ln (n!)} \\ a_{n} = \frac{1}{\ln (n!)} \\ b_{n} = \frac{1}{\ln (n^{n})} = \frac{1}{n \cdot \ln (n)} \\ b_{2^{n}} = \frac{1}{2^{n} \cdot n \cdot \ln (2)} \\ \sum_{n = 1}^{\infty} \frac{2^{n}}{2^{n} \cdot n \cdot \ln (2)} = \frac{1}{\ln (2)} \sum_{n = 1}^{\infty} \frac{1}{n} diverges \\ ⟹ By the Cauchy’s condensation test: \sum_{n = 1}^{\infty} b_{n} diverges \\ a_{n} > b_{n} ⟹ By the direct comparison test: \sum_{n = 1}^{\infty} a_{n} diverges \\ \frac{1}{\ln (n!)} is monotonically non-increasing and lim_{n \to \infty} \frac{1}{\ln (n!)} = 0 \\ ⟹ By the Leibniz criterion: \sum_{n = 1}^{\infty} \frac{(- 1)^{n}}{\ln (n!)} converges \\ ⟹ \sum_{n = 1}^{\infty} \frac{(- 1)^{n}}{\ln (n!)} converges conditionally \end{matrix}

2e

\begin{matrix} \sum_{n = 1}^{\infty} \frac{(- 1)^{n}}{2^{n}} \cdot {(1 - \cos (\frac{2}{n}))}^{n} \\ Solution: \\ Let a_{n} = \frac{(- 1)^{n} {(1 - \cos (\frac{2}{n}))}^{n}}{2^{n}} \\ lim_{n \to \infty} \sqrt[n]{| a_{n} |} = lim_{n \to \infty} \frac{(1 - \cos (\frac{2}{n}))}{2} \\ n \geq 2 ⟹ 0 < \frac{2}{n} < \frac{π}{2} ⟹ 0 < \cos (\frac{2}{n}) < 1 \\ ⟹ \frac{(1 - \cos (\frac{2}{n}))}{2} < \frac{1}{2} \\ ⟹ lim_{n \to \infty} \frac{(1 - \cos (\frac{2}{n}))}{2} \leq \frac{1}{2} < 1 \\ ⟹ By the root test: \sum_{n = 1}^{\infty} \frac{(- 1)^{n}}{2^{n}} \cdot {(1 - \cos (\frac{2}{n}))}^{n} converges absolutely \end{matrix}

2f

\begin{matrix} \sum_{n = 1}^{\infty} \frac{\sin^{2} (\frac{n}{2})}{n} \\ Solution: \\ \cos (n) = 1 - 2 \sin^{2} (\frac{n}{2}) \\ \sin^{2} (\frac{n}{2}) = \frac{1 - \cos (n)}{2} \\ ⟹ \sum_{n = 1}^{\infty} \frac{\sin^{2} (\frac{n}{2})}{n} = \sum_{n = 1}^{\infty} \frac{1 - \cos (n)}{2 n} = \frac{1}{2} \cdot (\sum_{n = 1}^{\infty} \frac{1}{n} - \sum_{n = 1}^{\infty} \frac{\cos (n)}{n}) \\ \sum_{n = 1}^{\infty} \cos (n) is bounded, \frac{1}{n} is monotonically decreasing and \frac{1}{n} \to 0 \\ ⟹ By the Dirichlet’s test: \sum_{n = 1}^{\infty} \frac{\cos (n)}{n} converges \\ \sum_{n = 1}^{\infty} \frac{1}{n} diverges \\ ⟹ By the series arithmetics: \sum_{n = 1}^{\infty} \frac{1}{n} - \sum_{n = 1}^{\infty} \frac{\cos (n)}{n} diverges \\ ⟹ \sum_{n = 1}^{\infty} \frac{\sin^{2} (\frac{n}{2})}{n} diverges \end{matrix}

2g

\begin{matrix} \sum_{n = 1}^{\infty} (- 1)^{n} {(\frac{n}{n + 1})}^{n^{2}} \\ Solution: \\ lim_{n \to \infty} \sqrt[n]{| (- 1)^{n} {(\frac{n}{n + 1})}^{n^{2}} |} = lim_{n \to \infty} {(\frac{n}{n + 1})}^{n} = lim_{n \to \infty} \frac{1}{{(1 + \frac{1}{n})}^{n}} = \frac{1}{e} < 1 \\ ⟹ By the root test: \sum_{n = 1}^{\infty} (- 1)^{n} {(\frac{n}{n + 1})}^{n^{2}} converges absolutely \end{matrix}

2h

\begin{matrix} \sum_{n = 1}^{\infty} \frac{n^{n}}{(n + 1)^{n + 1}} \\ Solution: \\ Let a_{n} = \frac{n^{n}}{(n + 1)^{n + 1}} \\ Let b_{n} = \frac{1}{n} \\ lim_{n \to \infty} \frac{a_{n}}{b_{n}} = lim_{n \to \infty} \frac{n^{n + 1}}{(n + 1)^{n + 1}} = lim_{n \to \infty} {(1 - \frac{1}{n + 1})}^{n + 1} = \frac{1}{e} \\ \sum_{n = 1}^{\infty} b_{n} diverges \\ ⟹ By the limit comparison test: \sum_{n = 1}^{\infty} a_{n} diverges \\ ⟹ \sum_{n = 1}^{\infty} \frac{n^{n}}{(n + 1)^{n + 1}} diverges \end{matrix}

2i

\begin{matrix} \sum_{n = 1}^{\infty} (- 1)^{n} (\sqrt{n + 5} - \sqrt{n + 4}) \\ Solution: \\ (- 1)^{n} (\sqrt{n + 5} - \sqrt{n + 4}) = \frac{(- 1)^{n} (n + 5 - (n + 4))}{\sqrt{n + 5} + \sqrt{n + 4}} = \frac{(- 1)^{n}}{\sqrt{n + 5} + \sqrt{n + 4}} \\ \frac{1}{\sqrt{n + 5} + \sqrt{n + 4}} > \frac{1}{\sqrt{2 n} + \sqrt{2 n}} = \frac{1}{2 \sqrt{2 n}} \\ \sum_{n = 1}^{\infty} \frac{1}{2 \sqrt{2 n}} diverges \\ ⟹ By the direct comparison test: \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n + 5} + \sqrt{n + 4}} diverges \\ \frac{1}{\sqrt{n + 5} + \sqrt{n + 4}} is monotonically decreasing and \frac{1}{\sqrt{n + 5} + \sqrt{n + 4}} \to 0 \\ ⟹ By the Leibniz criterion: \sum_{n = 1}^{\infty} \frac{(- 1)^{n}}{\sqrt{n + 5} + \sqrt{n + 4}} converges \\ ⟹ \sum_{n = 1}^{\infty} (- 1)^{n} (\sqrt{n + 5} - \sqrt{n + 4}) converges conditionally \end{matrix}

3

\begin{matrix} \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{(2 n - 1)^{α}} \\ Determine for which values of α series converges absolutely, conditionally or diverges \\ Solution: \\ \sum_{n = 1}^{\infty} | \frac{(- 1)^{n + 1}}{(2 n - 1)^{α}} | = \sum_{n = 1}^{\infty} \frac{1}{(2 n - 1)^{α}} \\ lim_{n \to \infty} \frac{\frac{1}{(2 n - 1)^{α}}}{\frac{1}{n^{α}}} = lim_{n \to \infty} {(\frac{n}{2 n - 1})}^{α} = lim_{n \to \infty} {(\frac{1}{2 - \frac{1}{n}})}^{α} = \frac{1}{2^{α}} \\ ⟹ By the limit comparison test: \sum_{n = 1}^{\infty} \frac{1}{(2 n - 1)^{α}} converges ⟺ \sum_{n = 1}^{\infty} \frac{1}{n^{α}} converges \\ ⟺ α > 1 \\ ⟹ \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{(2 n - 1)^{α}} converges absolutely when α > 1 \\ \frac{1}{(2 n - 1)^{α}} is monotonically decreasing and \frac{1}{(2 n - 1)^{α}} \to 0 when α > 0 \\ ⟹ By the Leibniz criterion: \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{(2 n - 1)^{α}} converges when α > 0 \\ Let α \leq 0 \\ Then ∄ lim_{n \to \infty} \frac{(- 1)^{n}}{(2 n - 1)^{α}} ⟹ \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{(2 n - 1)^{α}} diverges \\ ⟹ {\begin{cases} \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{(2 n - 1)^{α}} converges absolutely & α > 1 \\ \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{(2 n - 1)^{α}} converges conditionally & 0 < α \leq 1 \\ \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{(2 n - 1)^{α}} diverges & α \leq 0 \end{cases} \end{matrix}

4a

\begin{matrix} \sum_{n = 1}^{\infty} a_{n} > 0 \\ \exists p : by the limit comparison test with \frac{1}{n^{p}}, \sum_{n = 1}^{\infty} a_{n} diverges \\ Show that it can be proved that the limit comparison test with \frac{1}{n} \\ will also show divergence \\ Solution: \\ Let lim_{n \to \infty} \frac{a_{n}}{\frac{1}{n^{p}}} = L \\ By the limit comparison test \sum_{n = 1}^{\infty} a_{n} diverges \\ ⟹ \sum_{n = 1}^{\infty} \frac{1}{n^{p}} also diverges \\ ⟹ L must be > 0 or \infty, otherwise test won’t show divergence of a_{n} \\ If L > 0 then \sum_{n = 1}^{\infty} \frac{1}{n^{p}} diverges by limit comparison test \\ If L = \infty then \sum_{n = 1}^{\infty} \frac{1}{n^{p}} must diverge \\ otherwise limit comparison test won’t show divergence \\ \sum_{n = 1}^{\infty} \frac{1}{n^{p}} diverges ⟹ p \leq 1 \\ p \leq 1 ⟹ n^{p} \leq n ⟹ \frac{1}{n^{p}} \geq \frac{1}{n} \\ ⟹ \frac{a_{n}}{\frac{1}{n^{p}}} \leq \frac{a_{n}}{\frac{1}{n}} ⟹ L = lim_{n \to \infty} \frac{a_{n}}{\frac{1}{n^{p}}} \leq lim_{n \to \infty} \frac{a_{n}}{\frac{1}{n}} \\ ⟹ lim_{n \to \infty} \frac{a_{n}}{\frac{1}{n}} \geq L > 0 \\ ⟹ Limit comparison test with \frac{1}{n} will also show divergence of \sum_{n = 1}^{\infty} a_{n} \end{matrix}

4b

\begin{matrix} \sum_{n = 1}^{\infty} a_{n} > 0 \\ \exists p : by the limit comparison test with \frac{1}{n^{p}}, \sum_{n = 1}^{\infty} a_{n} converges \\ Show that it can be proved that the limit comparison test with \frac{1}{n^{q}}, 1 < q < p \\ will also show convergence \\ Solution: \\ Let lim_{n \to \infty} \frac{a_{n}}{\frac{1}{n^{p}}} = L \\ By the limit comparison test \sum_{n = 1}^{\infty} a_{n} converges \\ ⟹ \sum_{n = 1}^{\infty} \frac{1}{n^{p}} also converges \\ ⟹ L must be \geq 0 and L \neq \infty, otherwise test won’t show convergence of a_{n} \\ If L = 0, then \sum_{n = 1}^{\infty} \frac{1}{n^{p}} converges \\ If L > 0 and L \neq \infty then \sum_{n = 1}^{\infty} must converge \\ otherwise limit comparison test won’t show convergence \\ \sum_{n = 1}^{\infty} \frac{1}{n^{p}} converges ⟹ p > 1 \\ p > 1 \land p \in R ⟹ \exists q = \frac{p + 1}{2}, 1 < q < p \\ 1 < q < p ⟹ n^{q} < n^{p} ⟹ \frac{1}{n^{q}} > \frac{1}{n^{p}} \\ ⟹ \frac{a_{n}}{\frac{1}{n^{q}}} < \frac{a_{n}}{\frac{1}{n^{p}}} ⟹ lim_{n \to \infty} \frac{a_{n}}{\frac{1}{n^{q}}} \leq lim_{n \to \infty} \frac{a_{n}}{\frac{1}{n^{p}}} \\ ⟹ lim_{n \to \infty} \frac{a_{n}}{\frac{1}{n^{q}}} \geq 0 and \neq \infty \\ ⟹ Limit comparison test with \frac{1}{n^{q}} will also show convergence of \sum_{n = 1}^{\infty} a_{n} \end{matrix}

4c

\begin{matrix} Give an example of divergent series, whose divergence cannot be proved \\ by the limit comparison test with \frac{1}{n^{p}} \\ Solution: \\ Let a_{n} = \frac{1}{n \ln n} \\ \sum_{n = 1}^{\infty} a_{n} diverges by the Cauchy’s condensation test \\ If the limit comparison test with \frac{1}{n^{p}} will show divergence \\ then the limit comparison test with \frac{1}{n} will also show divergence \\ lim_{n \to \infty} \frac{a_{n}}{\frac{1}{n}} = lim_{n \to \infty} \frac{1}{\ln n} = 0 \\ Which does not show divergence \\ ⟹ Limit comparison test with \frac{1}{n^{p}} also doesn’t show divergence \end{matrix}

4c

\begin{matrix} Give an example of convergent series, whose divergence cannot be proved \\ by the limit comparison test with \frac{1}{n^{p}} \\ Solution: \\ Let a_{n} = \frac{1}{n (\ln n)^{2}} \\ \sum_{n = 1}^{\infty} a_{n} converges by Cauchy’s condensation test \\ lim_{n \to \infty} \frac{a_{n}}{\frac{1}{n^{p}}} = lim_{n \to \infty} \frac{n^{p - 1}}{(\ln n)^{2}} = lim_{n \to \infty} {(\frac{n^{(p - 1) / 2}}{\ln n})}^{2} \underset{p > 1 ⟹ \frac{p - 1}{2} > 0}{=} \infty \\ Which does not show convergence \\ ⟹ Limit comparison test with \frac{1}{n^{p}} doesn’t show convergence \end{matrix}