Infi-1 8

1

\begin{matrix} \sum_{n = 1}^{\infty} \frac{n^{4} - n^{2} + 6}{2 n^{3} - n + 4} \\ Determine whether series converge or diverge \\ Solution: \\ a_{n} = \frac{n^{4} - n^{2} + 6}{2 n^{3} - n + 4} = \frac{n - \overset{\to 0}{\overset{⏞}{\frac{1}{n}}} + \overset{\to 0}{\overset{⏞}{\frac{6}{n^{3}}}}}{2 - \underset{\to 0}{\underset{⏟}{\frac{1}{n^{2}}}} + \underset{\to 0}{\underset{⏟}{\frac{4}{n^{3}}}}} \\ ⟹ lim_{n \to \infty} a_{n} = lim_{n \to \infty} \frac{n}{2} = \infty \neq 0 \\ ⟹ \sum_{n = 1}^{\infty} a_{n} diverge \end{matrix}

2

\begin{matrix} \sum_{n = 1}^{\infty} \frac{4}{n \cdot 2^{\ln n}} \\ Determine whether series converge or diverge \\ Solution: \\ 2^{\ln n} = e^{\ln (2) \cdot \ln n} = e^{\ln n \cdot \ln (2)} = n^{\ln (2)} \\ ⟹ \sum_{n = 1}^{\infty} \frac{4}{n \cdot 2^{\ln n}} = 4 \sum_{n = 1}^{\infty} \frac{1}{n^{1 + \ln (2)}} \\ 1 + \ln (2) > 1 ⟹ \sum_{n = 1}^{\infty} \frac{1}{n^{1 + \ln (2)}} converge \\ ⟹ \sum_{n = 1}^{\infty} \frac{4}{n \cdot 2^{\ln n}} converge \end{matrix}

3

\begin{matrix} \sum_{n = 1}^{\infty} \frac{5 n^{6}}{2^{n}} \\ Determine whether series converge or diverge \\ Solution: \\ \sqrt[n]{\frac{5 n^{6}}{2^{n}}} = \frac{\sqrt[n]{5 n^{6}}}{2} = \frac{\underset{\to 1}{\underset{⏟}{\sqrt[n]{5}}} (\underset{\to 1}{\underset{⏟}{\sqrt[n]{n}}})^{6}}{2} \to \frac{1 \cdot 1^{6}}{2} = \frac{1}{2} < 1 \\ ⟹ By the root test, \sum_{n = 1}^{\infty} \frac{5 n^{6}}{2^{n}} converge \end{matrix}

4

\begin{matrix} \sum_{n = 2}^{\infty} \frac{\sin^{4} (\frac{n π}{7}) + \cos^{2} (\frac{2 n π}{9})}{n^{2} - n} \\ Determine whether series converge or diverge \\ Solution: \\ \frac{\sin^{4} (\frac{n π}{7}) + \cos^{2} (\frac{2 n π}{9})}{n^{2} - n} \leq \frac{1 + 1}{n^{2} - n} = \frac{2}{n^{2} - n} \\ Let a_{n} = \frac{2}{n^{2} - n} \geq 0 \\ Let b_{n} = \frac{1}{n^{2}} \geq 0 \\ \frac{a_{n}}{b_{n}} = \frac{\frac{1}{n^{2} - n}}{\frac{1}{n^{2}}} = \frac{n^{2}}{n^{2} - n} = \frac{1}{1 - \frac{1}{n}} \to 1 \\ lim_{n \to \infty} \frac{a_{n}}{b_{n}} = 1 > 0 \\ ⟹ By the limit comparison test: \sum_{n = 1}^{\infty} a_{n} converge ⟺ \sum_{n = 1}^{\infty} b_{n} converge \\ \sum_{n = 1}^{\infty} b_{n} = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} converge \\ ⟹ \sum_{n = 1}^{\infty} a_{n} = \sum_{n = 1}^{\infty} \frac{2}{n^{2} - n} converge ⟹ \sum_{n = 2}^{\infty} \frac{2}{n^{2} - n} converge \\ \frac{\sin^{4} (\frac{n π}{7}) + \cos^{2} (\frac{2 n π}{9})}{n^{2} - n} \leq \frac{2}{n^{2} - n} \\ ⟹ By the direct comparison test: \\ \sum_{n = 2}^{\infty} \frac{2}{n^{2} - n} converge ⟹ \sum_{n = 2}^{\infty} \frac{\sin^{4} (\frac{n π}{7}) + \cos^{2} (\frac{2 n π}{9})}{n^{2} - n} converge \end{matrix}

5

\begin{matrix} \sum_{n = 1}^{\infty} \frac{1}{(\ln (n + 3))^{n}} \\ Determine whether series converge or diverge \\ Solution: \\ Let a_{n} = \frac{1}{(\ln (n + 3))^{n}} \\ lim_{n \to \infty} \sqrt[n]{| a_{n} |} = lim_{n \to \infty} \sqrt[n]{\frac{1}{(\ln (n + 3))^{n}}} = lim_{n \to \infty} \frac{1}{\ln (n + 3)} = 0 \\ lim_{n \to \infty} \sqrt[n]{| a_{n} |} = 0 < 1 \\ ⟹ By the root test: \sum_{n = 1}^{\infty} \frac{1}{(\ln (n + 3))^{n}} converge \end{matrix}

6

\begin{matrix} \sum_{n = 1}^{\infty} \frac{2 n + 5}{(n + 2)^{2} (6 n - 5)^{3}} \\ Determine whether series converge or diverge \\ Solution: \\ Let a_{n} = \frac{2 n + 5}{(n + 2)^{2} (6 n - 5)^{3}} \\ Let b_{n} = \frac{1}{n^{4}} \\ \frac{a_{n}}{b_{n}} = \frac{(2 n + 5) n^{4}}{(n + 2)^{2} (6 n - 5)^{3}} = \frac{n^{5} (2 + \frac{5}{n})}{n^{2} {(1 + \frac{2}{n})}^{2} n^{3} {(6 - \frac{5}{n})}^{3}} = \frac{2 + \frac{5}{n}}{{(1 + \frac{2}{n})}^{2} {(6 - \frac{5}{n})}^{3}} \to \frac{2}{6^{3}} = \frac{1}{108} \\ ⟹ lim_{n \to \infty} \frac{a_{n}}{b_{n}} = \frac{1}{108} < 1 \\ \sum_{n = 1}^{\infty} b_{n} = \sum_{n = 1}^{\infty} \frac{1}{n^{4}} converge \\ ⟹ By the limit comparison test: \sum_{n = 1}^{\infty} \frac{2 n + 5}{(n + 2)^{2} (6 n - 5)^{3}} converge \end{matrix}

7

\begin{matrix} \sum_{n = 1}^{\infty} \frac{1}{n (\ln n)^{2}} \\ Determine whether series converge or diverge \\ Solution: \\ Let a_{n} = \frac{1}{n (\ln n)^{2}} \\ a_{2^{n}} = \frac{1}{2^{n} (\ln (2^{n}))^{2}} = \frac{1}{2^{n} (n \ln (2))^{2}} = \frac{1}{2^{n} n^{2} (\ln (2))^{2}} \\ \sum_{n = 1}^{\infty} \frac{2^{n}}{2^{n} n^{2} (\ln (2))^{2}} = \frac{1}{(\ln (2))^{2}} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} \\ \sum_{n = 1}^{\infty} \frac{1}{n^{2}} converge ⟹ \sum_{n = 1}^{\infty} 2^{n} a_{2^{n}} converge \\ a_{n} \geq 0 \\ n is monotonically increasing \\ (\ln n)^{2} is monotonically increasing \\ ⟹ n (\ln n)^{2} is monotonically increasing \\ ⟹ a_{n} is monotonically decreasing \\ ⟹ By the condensation test: \sum_{n = 1}^{\infty} a_{n} converge \\ ⟹ \sum_{n = 1}^{\infty} \frac{1}{n (\ln n)^{2}} converge \end{matrix}

8

\begin{matrix} \sum_{n = 1}^{\infty} \frac{n^{6}}{3^{n^{2}}} \\ Determine whether series converge or diverge \\ Solution: \\ Let a_{n} = \frac{n^{6}}{3^{n^{2}}} \\ lim_{n \to \infty} \sqrt[n]{| a_{n} |} = lim_{n \to \infty} \sqrt[n]{\frac{n^{6}}{3^{n^{2}}}} = lim_{n \to \infty} \frac{(\overset{\to 1}{\overset{⏞}{\sqrt[n]{n}}})^{6}}{\underset{\to \infty}{\underset{⏟}{3^{n}}}} = 0 \\ lim_{n \to \infty} \sqrt[n]{a_{n}} = 0 < 1 \\ ⟹ By the root test: \sum_{n = 1}^{\infty} \frac{n^{6}}{3^{n^{2}}} converge \end{matrix}

9

\begin{matrix} \sum_{n = 1}^{\infty} \frac{1 - \sin^{2} (n^{6} + n^{4} + n^{2} + 1)}{n \sqrt{n}} \\ Determine whether series converge or diverge \\ Solution: \\ 0 \leq \sin^{2} (n^{6} + n^{4} + n^{2} + 1) \leq 1 \\ ⟹ 0 \leq 1 - \sin^{2} (n^{6} + n^{4} + n^{2} + 1) \leq 1 \\ ⟹ 0 \leq \frac{1 - \sin^{2} (n^{6} + n^{4} + n^{2} + 1)}{n \sqrt{n}} \leq \frac{1}{n \sqrt{n}} \\ \sum_{n = 1}^{\infty} \frac{1}{n \sqrt{n}} = \sum_{n = 1}^{\infty} \frac{1}{n^{3 / 2}} converge \\ ⟹ By the direct comparison test: \sum_{n = 1}^{\infty} \frac{1 - \sin^{2} (n^{6} + n^{4} + n^{2} + 1)}{n \sqrt{n}} converge \end{matrix}

10

\begin{matrix} \sum_{n = 1}^{\infty} \frac{8^{n} \cdot n!}{n^{n}} \\ Determine whether series converge or diverge \\ Solution: \\ Let a_{n} = \frac{8^{n} \cdot n!}{n^{n}} \\ \frac{a_{n + 1}}{a_{n}} = \frac{8^{n + 1} \cdot (n + 1)! \cdot n^{n}}{(n + 1)^{n + 1} \cdot 8^{n} \cdot n!} = 8 (n + 1) \cdot \frac{n^{n}}{(n + 1)^{n + 1}} = 8 {(\frac{n}{n + 1})}^{n} = \\ = \frac{8}{{(\frac{n + 1}{n})}^{n}} = \frac{8}{{(1 + \frac{1}{n})}^{n}} \to \frac{8}{e} \\ lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = \frac{8}{e} > 1 \\ ⟹ By the ratio test: \sum_{n = 1}^{\infty} \frac{8^{n} \cdot n!}{n^{n}} diverge \end{matrix}

11

\begin{matrix} \sum_{n = 1}^{\infty} \frac{(1 + \sqrt{2}) (1 + \sqrt{3}) \dots (1 + \sqrt{n})}{n!} \\ Determine whether series converge or diverge \\ Solution: \\ Let a_{n} = \frac{(1 + \sqrt{2}) (1 + \sqrt{3}) \dots (1 + \sqrt{n})}{n!} \\ \frac{a_{n + 1}}{a_{n}} = \frac{(1 + \sqrt{2}) (1 + \sqrt{3}) \dots (1 + \sqrt{n + 1}) \cdot n!}{(n + 1)! \cdot (1 + \sqrt{2}) (1 + \sqrt{3}) \dots (1 + \sqrt{n})} = \frac{1 + \sqrt{n + 1}}{n + 1} = \\ = \frac{\frac{1}{n} + \sqrt{\frac{1}{n} + \frac{1}{n^{2}}}}{1 + \frac{1}{n}} \to \frac{0}{1} = 0 \\ lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = 0 < 1 \\ ⟹ By the ratio test: \sum_{n = 1}^{\infty} \frac{(1 + \sqrt{2}) (1 + \sqrt{3}) \dots (1 + \sqrt{n})}{n!} converge \end{matrix}

12

\begin{matrix} \sum_{n = 1}^{\infty} \frac{n^{2} + \sqrt{n^{2} + n} + \sqrt{n^{5} - n}}{\sqrt[3]{n^{8} - 5} \sqrt{n^{4} + n + 1}} \\ Determine whether series converge or diverge \\ Solution: \\ Let a_{n} = \frac{n^{2} + \sqrt{n^{2} + n} + \sqrt{n^{5} - n}}{\sqrt[3]{n^{8} - 5} \sqrt{n^{4} + n + 1}} \\ Dominant exponents are: \frac{5}{2} - \frac{14}{3} = \frac{15 - 28}{6} = - \frac{13}{6} \\ Let b_{n} = \frac{1}{\sqrt[6]{n^{13}}} \\ \frac{a_{n}}{b_{n}} = \frac{(n^{2} + \sqrt{n^{2} + n} + \sqrt{n^{5} - n}) \sqrt[6]{n^{13}}}{\sqrt[3]{n^{8} - 5} \sqrt{n^{4} + n + 1}} = \\ = \frac{n^{25 / 6} + \sqrt{n^{38 / 6} + n^{32 / 6}} + \sqrt{n^{56 / 6} - n^{32 / 6}}}{\sqrt[3]{n^{8} - 5} \sqrt{n^{4} + n + 1}} = \frac{\frac{n^{25 / 6}}{n^{28 / 6}} + \frac{\sqrt{n^{38 / 6} + n^{32 / 6}}}{n^{28 / 6}} + \frac{\sqrt{n^{56 / 6} - n^{32 / 6}}}{n^{28 / 6}}}{\frac{\sqrt[3]{n^{8} - 5}}{n^{16 / 6}} \frac{\sqrt{n^{4} + n + 1}}{n^{2}}} = \\ = \frac{\overset{\to 0}{\overset{⏞}{\frac{1}{\sqrt{n}}}} + \overset{\to 0}{\overset{⏞}{\sqrt{\frac{1}{n^{3}} + \frac{1}{n^{4}}}}} + \overset{\to 0}{\overset{⏞}{\sqrt{1 - \frac{1}{n^{4}}}}}}{\underset{\to \sqrt{1} = 1}{\underset{⏟}{\sqrt[3]{1 - \frac{5}{n^{8}}}}} \underset{\to \sqrt{1} = 1}{\underset{⏟}{\sqrt{1 + \frac{1}{n^{3}} + \frac{1}{n^{4}}}}}} \to \frac{0}{1 \cdot 1} = 0 \\ lim_{n \to \infty} \frac{a_{n}}{b_{n}} = 0 < 1 \\ \sum_{n = 1}^{\infty} b_{n} = \sum_{n = 1}^{\infty} \frac{1}{n^{13 / 6}} converge \\ ⟹ By the limit comparison test: \sum_{n = 1}^{\infty} \frac{n^{2} + \sqrt{n^{2} + n} + \sqrt{n^{5} - n}}{\sqrt[3]{n^{8} - 5} \sqrt{n^{4} + n + 1}} converge \end{matrix}