Infi-1 7

1a

\begin{matrix} a_{n} = {(1 - \frac{1}{n^{2} + 6})}^{2 n^{2} + n + 5} \\ Find lim_{n \to \infty} a_{n} \\ Solution: \\ a_{n} = {(\underset{\to 1}{\underset{⏟}{1 - \frac{1}{n^{2} + 6}}})}^{\underset{\to \infty}{\underset{⏟}{2 n^{2} + n + 5}}} \\ ⟹ lim_{n \to \infty} a_{n} = e^{lim_{n \to \infty} - (2 n^{2} + n + 5) / (n^{2} + 6)} \\ lim_{n \to \infty} - \frac{2 n^{2} + n + 5}{n^{2} + 6} = - lim_{n \to \infty} \frac{2 + \overset{\to 0}{\overset{⏞}{\frac{1}{n}}} + \overset{\to 0}{\overset{⏞}{\frac{5}{n^{2}}}}}{1 + \underset{\to 0}{\underset{⏟}{\frac{6}{n^{2}}}}} = - 2 \\ ⟹ lim_{n \to \infty} a_{n} = e^{- 2} = \frac{1}{e^{2}} \end{matrix}

1b

\begin{matrix} b_{n} = {(\frac{n^{5} - 3 n}{n^{5} + 2 n^{2} + n})}^{n^{2} - 16} \\ Find lim_{n \to \infty} b_{n} \\ Solution: \\ b_{n} = {(\frac{n^{5} - 3 n}{n^{5} + 2 n^{2} + n})}^{n^{2} - 16} = {(\frac{n^{5} + 2 n^{2} + n - 2 n^{2} - 4 n}{n^{5} + 2 n^{2} + n})}^{n^{2} - 16} = \\ = {(1 - \frac{2 n^{2} + 4 n}{n^{5} + 2 n^{2} + n})}^{n^{2} - 16} = {(\underset{\to 1}{\underset{⏟}{1 - \frac{2 + \frac{4}{n}}{n^{3} + 2 + \frac{1}{n}}}})}^{\underset{\to \infty}{\underset{⏟}{n^{2} - 16}}} \\ ⟹ lim_{n \to \infty} b_{n} = e^{lim_{n \to \infty} - (n^{2} - 16) (2 n^{2} + 4 n) / (n^{5} + 2 n^{2} + n)} \\ lim_{n \to \infty} - \frac{(n^{2} - 16) (2 n^{2} + 4 n)}{n^{5} + 2 n^{2} + n} = - lim_{n \to \infty} \frac{2 n^{4} + 4 n^{3} - 32 n^{2} - 64 n}{n^{5} + 2 n^{2} + n} = \\ = - lim_{n \to \infty} \underset{\to 0}{\underset{⏟}{\frac{1}{n}}} \cdot \underset{\to 2}{\underset{⏟}{\frac{2 + \frac{4}{n} - \frac{32}{n^{2}} - \frac{64}{n^{3}}}{1 + \frac{2}{n^{3}} + \frac{1}{n^{4}}}}} = 0 \\ ⟹ lim_{n \to \infty} b_{n} = e^{0} = 1 \end{matrix}

1c

\begin{matrix} c_{n} = {(\frac{n^{5} - 1}{n^{5} + 2 n})}^{n^{5} + 3} \\ Find lim_{n \to \infty} c_{n} \\ Solution: \\ c_{n} = {(\frac{n^{5} + 2 n - 2 n - 1}{n^{5} + 2 n})}^{n^{5} + 3} = {(1 - \frac{2 n + 1}{n^{5} + 2 n})}^{n^{5} + 3} = {(\underset{\to 1}{\underset{⏟}{1 - \frac{2 + \frac{1}{n}}{n^{4} + 2}}})}^{\underset{\to \infty}{\underset{⏟}{n^{5} + 3}}} \\ ⟹ lim_{n \to \infty} c_{n} = e^{lim_{n \to \infty} - (n^{5} + 3) (2 n + 1) / (n^{5} + 2 n)} \\ lim_{n \to \infty} - \frac{(n^{5} + 3) (2 n + 1)}{n^{5} + 2 n} = - lim_{n \to \infty} \frac{2 n^{6} + n^{5} + 6 n + 3}{n^{5} + 2 n} \\ = - lim_{n \to \infty} \underset{\to \infty}{\underset{⏟}{n}} \cdot \underset{\to 2}{\underset{⏟}{\frac{2 + \frac{1}{n} + \frac{6}{n^{5}} + \frac{3}{n^{6}}}{1 + \frac{2}{n^{4}}}}} = - \infty \\ ⟹ lim_{n \to \infty} c_{n} = e^{- \infty} = \frac{1}{e^{\infty}} = 0 \end{matrix}

1d

\begin{matrix} d_{n} = {(\frac{n + 2}{n})}^{n \sqrt{n + 1}} \\ Find lim_{n \to \infty} d_{n} \\ Solution: \\ d_{n} = {(\underset{\to 1}{\underset{⏟}{1 + \frac{2}{n}}})}^{\underset{\to \infty}{\underset{⏟}{n \sqrt{n + 1}}}} \\ ⟹ lim_{n \to \infty} d_{n} = e^{lim_{n \to \infty} 2 \sqrt{n + 1}} = e^{\infty} = \infty \end{matrix}

1e

\begin{matrix} p_{n} = {(\frac{n \sqrt{n}}{n^{2} - n + 1})}^{16 n^{2} - 16 n - 3} \\ Find lim_{n \to \infty} p_{n} \\ Solution: \\ (p_{1})_{n} = \frac{n \sqrt{n}}{n^{2} - n + 1} = \frac{1}{\underset{\to \infty}{\underset{⏟}{\sqrt{n}}} - \underset{\to 0}{\underset{⏟}{\frac{1}{\sqrt{n}}}} + \underset{\to 0}{\underset{⏟}{\frac{1}{n \sqrt{n}}}}} \to \frac{1}{\infty} = 0 \\ (p_{2})_{n} = 16 n^{2} - 16 n - 3 \to \infty \\ ⟹ p_{n} = (p_{1})_{n}^{(p_{2})_{n}} \to 0^{\infty} = 0 \\ ⟹ lim_{n \to \infty} p_{n} = 0 \end{matrix}

1f

\begin{matrix} q_{n} = {(\frac{n^{2} + \ln (n)}{\ln (n) + n})}^{\ln (n)} \\ Find lim_{n \to \infty} q_{n} \\ Solution: \\ (q_{1})_{n} = \frac{n^{2} + \ln (n)}{n + \ln (n)} = \frac{\underset{\to \infty}{\underset{⏟}{n}} + \underset{\to 0}{\underset{⏟}{\frac{\ln (n)}{n}}}}{1 + \underset{\to 0}{\underset{⏟}{\frac{\ln (n)}{n}}}} \to \frac{\infty}{1} = \infty \\ (q_{2})_{n} = \ln (n) \to \infty \\ q_{n} = (q_{1})_{n}^{(q_{2})_{n}} \to \infty^{\infty} ⟹ lim_{n \to \infty} q_{n} = \infty \end{matrix}

2

\begin{matrix} a_{n} \to 1, b_{n} \to \infty \end{matrix}

2a

\begin{matrix} Give an example of a_{n}, b_{n} such that a_{n}^{b_{n}} \to 0 \\ Solution: \\ a_{n} = 1 - \frac{1}{n} \\ b_{n} = n^{2} \\ lim_{n \to \infty} a_{n}^{b_{n}} = e^{lim_{n \to \infty} b_{n} (a_{n} - 1)} = e^{lim_{n \to \infty} - n^{2} / n} = e^{- \infty} = 0 \\ lim_{n \to \infty} a_{n}^{b_{n}} = 0 \end{matrix}

2b

\begin{matrix} Give an example of a_{n}, b_{n} such that a_{n}^{b_{n}} \to L, 0 < L < 1 \\ Solution: \\ a_{n} = 1 - \frac{1}{n} \\ b_{n} = 2 n \\ lim_{n \to \infty} a_{n}^{b_{n}} = e^{lim_{n \to \infty} b_{n} (a_{n} - 1)} = e^{lim_{n \to \infty} - 2 n / n} = e^{- 2} = \frac{1}{e^{2}} \end{matrix}

2c

\begin{matrix} Give an example of a_{n}, b_{n} such that a_{n}^{b_{n}} \to 1 \\ Solution: \\ a_{n} = 1 + \frac{1}{n^{2}} \\ b_{n} = n \\ lim_{n \to \infty} a_{n}^{b_{n}} = e^{lim_{n \to \infty} b_{n} (a_{n} - 1)} = e^{lim_{n \to \infty} n / n^{2}} = e^{0} = 1 \end{matrix}

2d

\begin{matrix} Give an example of a_{n}, b_{n} such that a_{n}^{b_{n}} \to L, L > 1 \\ Solution: \\ a_{n} = 1 + \frac{1}{n} \\ b_{n} = n \\ lim_{n \to \infty} a_{n}^{b_{n}} = e^{lim_{n \to \infty} b_{n} (a_{n} - 1)} = e^{lim_{n \to \infty} n / n} = e \end{matrix}

2e

\begin{matrix} Give an example of a_{n}, b_{n} such that a_{n}^{b_{n}} \to \infty \\ Solution: \\ a_{n} = 1 + \frac{1}{n} \\ b_{n} = n^{2} \\ lim_{n \to \infty} a_{n}^{b_{n}} = e^{lim_{n \to \infty} b_{n} (a_{n} - 1)} = e^{lim_{n \to \infty} n^{2} / n} = e^{\infty} = \infty \end{matrix}

2f

\begin{matrix} Give an example of a_{n}, b_{n} such that ∄ lim_{n \to \infty} a_{n}^{b_{n}} \\ Solution: \\ a_{n} = 1 + \frac{(- 1)^{n}}{n} \\ b_{n} = n \\ lim_{n \to \infty} a_{n}^{b_{n}} = e^{lim_{n \to \infty} b_{n} (a_{n} - 1)} = e^{lim_{n \to \infty} (- 1)^{n} n / n} = e^{lim_{n \to \infty} (- 1)^{n}} \\ ∄ lim_{n \to \infty} (- 1)^{n} ⟹ ∄ lim_{n \to \infty} a_{n}^{b_{n}} \end{matrix}

3a

\begin{matrix} Prove: \sum_{n = 1}^{\infty} \frac{e^{n}}{n^{6}} is divergent \\ Proof: \\ \frac{e^{n}}{n^{6}} \to \infty ⟹ \frac{e^{n}}{n^{6}} ↛ 0 \\ ⟹ \sum_{n = 1}^{\infty} \frac{e^{n}}{n^{6}} is divergent \end{matrix}

3b

\begin{matrix} Prove: \sum_{n = 1}^{\infty} \frac{n^{4} - 2 n + 3 e}{2 n^{4} + 12 n^{2} - n - π} is divergent \\ Proof: \\ a_{n} = \frac{n^{4} - 2 n + 3 e}{2 n^{4} + 12 n^{2} - n - π} = \frac{1 - \frac{2}{n^{3}} + \frac{3 e}{n^{4}}}{2 + \frac{12}{n^{2}} - \frac{1}{n^{3}} - \frac{π}{n^{4}}} \to \frac{1}{2} \\ ⟹ a_{n} ↛ 0 ⟹ \sum_{n = 1}^{\infty} a_{n} is divergent \end{matrix}

3c

\begin{matrix} Prove: \sum_{n = 1}^{\infty} \frac{n^{4}}{(n + 1) \sqrt{n + 2} \ln (n + 3)} is divergent \\ Proof: \\ a_{n} = \frac{n^{4}}{(n + 1) \sqrt{n + 2} \ln (n + 3)} = \underset{\to 1}{\underset{⏟}{\frac{n}{n + 1}}} \cdot \underset{\to \infty}{\underset{⏟}{\frac{n^{2}}{\sqrt{n + 2}}}} \cdot \underset{\to \infty}{\underset{⏟}{\frac{n}{\ln (n + 3)}}} \to \infty \\ ⟹ a_{n} ↛ 0 ⟹ \sum_{n = 1}^{\infty} a_{n} is divergent \end{matrix}

4a

\begin{matrix} Determine whether \sum_{n = 1}^{\infty} \sqrt{n + 1} - \sqrt{n} is convergent or divergent \\ And if convergent, find the value of the sum \\ Solution: \\ S_{1} = \sqrt{2} - \sqrt{1} \\ S_{2} = \sqrt{2} - \sqrt{1} + \sqrt{3} - \sqrt{2} \\ \dots \\ S_{N} = \sqrt{2} - \sqrt{1} + \sqrt{3} - \sqrt{2} + \dots + \sqrt{N} - \sqrt{N - 1} + \sqrt{N + 1} - \sqrt{N} \\ ⟹ S_{N} = \sqrt{N + 1} - \sqrt{1} \\ ⟹ S_{n} \to \infty ⟹ \sum_{n = 1}^{\infty} \sqrt{n + 1} - \sqrt{n} is divergent \end{matrix}

4b

\begin{matrix} Determine whether \sum_{n = 1}^{\infty} \frac{1}{4 n^{2} - 1} is convergent or divergent \\ And if convergent, find the value of the sum \\ Solution: \\ \frac{1}{4 n^{2} - 1} = \frac{1}{(2 n - 1) (2 n + 1)} = \frac{(2 n + 1) - (2 n - 1)}{2 (2 n + 1) (2 n - 1)} = \frac{1}{4 n - 2} - \frac{1}{4 n + 2} \\ ⟹ S_{1} = \frac{1}{4 - 2} - \frac{1}{4 + 2} = \frac{1}{2} - \frac{1}{6} \\ S_{2} = \frac{1}{4 - 2} - \frac{1}{4 + 2} + \frac{1}{8 - 2} - \frac{1}{8 + 2} = \frac{1}{2} - \frac{1}{6} + \frac{1}{6} - \frac{1}{10} \\ \dots \\ S_{N} = \frac{1}{2} - \frac{1}{6} + \frac{1}{6} - \frac{1}{10} + \dots + \frac{1}{4 N - 6} - \frac{1}{4 N - 2} + \frac{1}{4 N - 2} - \frac{1}{4 N + 2} \\ ⟹ S_{N} = \frac{1}{2} - \frac{1}{4 N + 2} \\ S_{n} = \frac{1}{2} - \frac{1}{4 n + 2} \to \frac{1}{2} \\ ⟹ \sum_{n = 1}^{\infty} \frac{1}{4 n^{2} - 1} = \frac{1}{2} \end{matrix}

5a

\begin{matrix} Determine whether \sum_{n = 0}^{\infty} \frac{2^{n} + (- 3)^{n}}{6^{n}} is convergent or divergent \\ And if convergent, find the value of the sum \\ Solution: \\ \frac{2^{n} + (- 3)^{n}}{6^{n}} = {(\frac{1}{3})}^{n} + {(- \frac{1}{2})}^{n} \\ ⟹ \sum_{n = 0}^{\infty} \frac{2^{n} + (- 3)^{n}}{6^{n}} = \underset{\frac{1}{3} < 1}{\underset{⏟}{\sum_{n = 0}^{\infty} {(\frac{1}{3})}^{n}}} + \underset{- \frac{1}{2} > - 1}{\underset{⏟}{\sum_{n = 0}^{\infty} {(- \frac{1}{2})}^{n}}} = \frac{1}{1 - \frac{1}{3}} + \frac{1}{1 + \frac{1}{2}} = \frac{3}{2} + \frac{2}{3} = \frac{13}{6} \\ ⟹ \sum_{n = 0}^{\infty} \frac{2^{n} + (- 3)^{n}}{6^{n}} = \frac{13}{6} \end{matrix}

5b

\begin{matrix} Determine whether \sum_{n = 0}^{\infty} \frac{2^{n} - 3^{n}}{6^{n}} is convergent or divergent \\ And if convergent, find the value of the sum \\ Solution: \\ \frac{2^{n} + 3^{n}}{6^{n}} = {(\frac{1}{3})}^{n} - {(\frac{1}{2})}^{n} \\ ⟹ \sum_{n = 0}^{\infty} \frac{2^{n} - 3^{n}}{6^{n}} = \underset{\frac{1}{3} < 1}{\underset{⏟}{\sum_{n = 0}^{\infty} {(\frac{1}{3})}^{n}}} - \underset{\frac{1}{2} < 1}{\underset{⏟}{\sum_{n = 0}^{\infty} {(\frac{1}{2})}^{n}}} = \frac{1}{1 - \frac{1}{3}} - \frac{1}{1 - \frac{1}{2}} = \frac{3}{2} - 2 = - \frac{1}{2} \\ ⟹ \sum_{n = 0}^{\infty} \frac{2^{n} - 3^{n}}{6^{n}} = - \frac{1}{2} \end{matrix}

5c

\begin{matrix} Determine whether \sum_{n = 1}^{\infty} \frac{2^{n} - 3^{n}}{6^{n}} is convergent or divergent \\ And if convergent, find the value of the sum \\ Solution: \\ \frac{2^{n} + 3^{n}}{6^{n}} = {(\frac{1}{3})}^{n} - {(\frac{1}{2})}^{n} \\ ⟹ \sum_{n = 1}^{\infty} \frac{2^{n} + (- 3)^{n}}{6^{n}} = \underset{\frac{1}{3} < 1}{\underset{⏟}{\sum_{n = 0}^{\infty} {(\frac{1}{3})}^{n}}} - \underset{\frac{1}{2} < 1}{\underset{⏟}{\sum_{n = 0}^{\infty} {(\frac{1}{2})}^{n}}} - \frac{2^{0} - 3^{0}}{6^{0}} = \\ = \frac{1}{1 - \frac{1}{3}} - \frac{1}{1 - \frac{1}{2}} - 0 = \frac{3}{2} - 2 = - \frac{1}{2} \\ ⟹ \sum_{n = 1}^{\infty} \frac{2^{n} - 3^{n}}{6^{n}} = - \frac{1}{2} \end{matrix}

5d

\begin{matrix} Determine whether \sum_{n = 1}^{\infty} \frac{3^{n} + 4^{n}}{n^{5}} is convergent or divergent \\ And if convergent, find the value of the sum \\ Solution: \\ a_{n} = \frac{3^{n} + 4^{n}}{n^{5}} = \underset{\to \infty}{\underset{⏟}{\frac{3^{n}}{n^{5}}}} + \underset{\to \infty}{\underset{⏟}{\frac{4^{n}}{n^{5}}}} \to \infty \\ ⟹ a_{n} ↛ 0 ⟹ \sum_{n = 1}^{\infty} a_{n} is divergent \end{matrix}

5e

\begin{matrix} Determine whether \sum_{n = 0}^{\infty} e^{1 - n} is convergent or divergent \\ And if convergent, find the value of the sum \\ Solution: \\ e^{1 - n} = e \cdot \frac{1}{e^{n}} = e \cdot {(\frac{1}{e})}^{n} \\ ⟹ \sum_{n = 0}^{\infty} e^{1 - n} = e \cdot \underset{\frac{1}{e} < 1}{\underset{⏟}{\sum_{n = 0}^{\infty} {(\frac{1}{e})}^{n}}} = e \cdot \frac{1}{1 - \frac{1}{e}} = \frac{e^{2}}{e - 1} \\ ⟹ \sum_{n = 0}^{\infty} e^{1 - n} = \frac{e^{2}}{e - 1} \end{matrix}

5f

\begin{matrix} Determine whether \sum_{n = 0}^{\infty} {(1 + \frac{1}{n})}^{n / e} is convergent or divergent \\ And if convergent, find the value of the sum \\ Solution: \\ a_{n} = {(1 + \frac{1}{n})}^{n / e} = {({(1 + \frac{1}{n})}^{n})}^{1 / e} \to e^{1 / e} \\ ⟹ a_{n} ↛ 0 ⟹ \sum_{n = 1}^{\infty} a_{n} is divergent \end{matrix}