Infi-1 5

1a

\begin{matrix} a > 0, b > 1 \\ Prove: \frac{b^{n}}{n^{a}} \to \infty \\ Proof: \\ Let x_{n} = \frac{b^{n}}{n^{a}} \\ | \frac{x_{n + 1}}{x_{n}} | = \frac{b^{n + 1} \cdot n^{a}}{b^{n} \cdot (n + 1)^{a}} = b \cdot {(\frac{n}{n + 1})}^{a} = b \cdot \underset{\to 1}{\underset{⏟}{{(\frac{1}{1 + \underset{\to 0}{\underset{⏟}{\frac{1}{n}}}})}^{a}}} \\ ⟹ lim_{n \to \infty} | \frac{x_{n + 1}}{x_{n}} | = b > 1 ⟹ lim_{n \to \infty} x_{n} = \infty \end{matrix}

1b

\begin{matrix} Prove: \frac{n}{\ln n} \to \infty \\ Proof: \\ \frac{n}{\ln n} = \frac{n}{2 \ln (\sqrt{n})} \\ \sqrt{n} \in R ⟹ \sqrt{n} \geq \ln (\sqrt{n}) \\ ⟹ \frac{n}{2 \ln (\sqrt{n})} = \frac{\sqrt{n}}{2} \cdot \frac{\sqrt{n}}{\ln (\sqrt{n})} \geq \frac{\sqrt{n}}{2} \to \infty \\ ⟹ lim_{n \to \infty} \frac{n}{\ln n} = \infty \end{matrix}

1c

\begin{matrix} a > 0 \\ Prove: \frac{n^{a}}{\ln n} \to \infty \\ Proof: \\ \ln n = \ln (n^{2 a / 2 a}) = \frac{2}{a} \ln (\sqrt{n^{a}}) \\ \frac{n^{a}}{\ln n} = \frac{\sqrt{n^{a}}}{\frac{2}{a}} \cdot \frac{\sqrt{n^{a}}}{\ln (\sqrt{n^{a}})} \\ \sqrt{n^{a}} \in R ⟹ \sqrt{n^{a}} \geq \ln (\sqrt{n^{a}}) \\ ⟹ \frac{\sqrt{n^{a}}}{\frac{2}{a}} \cdot \frac{\sqrt{n^{a}}}{\ln (\sqrt{n^{a}})} \geq \frac{a \sqrt{n^{a}}}{2} \to \infty \\ ⟹ lim_{n \to \infty} \frac{n^{a}}{\ln n} = \infty \end{matrix}

2a

\begin{matrix} a_{n} > 0 \\ \exists N : \forall n \geq N : a_{n + 1} > 2 a_{n} \\ Prove by definition: a_{n} \to \infty \\ Proof: \\ Let N_{1} > N \\ \forall n > N_{1} : a_{n} \geq 2 a_{N_{1}} > 4 a_{N_{1} - 1} > 8 a_{N_{1} - 2} > \dots > 2^{N_{1} - N + 1} a_{N} \\ Let M = 2^{N_{1} - N + 1} a_{N} \\ a_{N} > 0 ⟹ M > 0 \\ 2^{N_{1}} = \frac{M \cdot 2^{N - 1}}{a_{N}} ⟹ N_{1} = \log_{2} (M \cdot \frac{2^{N - 1}}{a_{N}}) \\ Let N_{1} = m a x (\log_{2} (M \cdot \frac{2^{N - 1}}{a_{N}}), N + 1) \\ ⟹ \forall M > 0 : \exists N_{1} > N : \forall n > N_{1} : a_{n} > M \\ ⟹ lim_{n \to \infty} a_{n} = \infty \end{matrix}

2b

\begin{matrix} a_{n} > 0 \\ Prove by definition: \frac{a_{n + 1}}{a_{n}} \to \infty ⟹ a_{n} \to \infty \\ Proof: \\ \frac{a_{n + 1}}{a_{n}} \to \infty ⟹ \forall M > 0 : \exists N_{1} : \forall n > N_{1} : \frac{a_{n + 1}}{a_{n}} > M ⟹ a_{n + 1} > M a_{n} \\ Let M = 2 \\ Let N = N_{1} + 1 \\ As proved in 2a, \exists N : \forall n \geq N : a_{n + 1} > 2 a_{n} ⟹ a_{n} \to \infty \\ ⟹ lim_{n \to \infty} a_{n} = \infty \end{matrix}

2c

\begin{matrix} a_{n} > 0 \\ Prove: \sqrt[n]{a_{n}} \to \infty ⟹ a_{n} \to \infty \\ Proof: \\ \sqrt[n]{a_{n}} \to \infty ⟹ \forall M > 0 : \exists N : \forall n > N : \sqrt[n]{a_{n}} > M ⟹ a_{n} > M^{n} \\ Let M > 1 ⟹ M^{n} \to \infty \\ ⟹ a_{n} \to \infty \end{matrix}

3a

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

3b

\begin{matrix} b_{n} = n \log (n) - 2^{n} \sqrt{n} \\ Find: lim_{n \to \infty} b_{n} \\ Solution: \\ - b_{n} = 2^{n} \sqrt{n} - n \log (n) \geq 2^{n} \sqrt{n} - \sqrt{n} \log (n) = \underset{\to \infty}{\underset{⏟}{\sqrt{n}}} \underset{> 0}{\underset{⏟}{(2^{n} - \log (n))}} \to \infty \\ ⟹ b_{n} \to - \infty \end{matrix}

3c

\begin{matrix} c_{n} = \sqrt{n + 16} - \sqrt{n + 9} \\ Find: lim_{n \to \infty} c_{n} \\ Solution: \\ c_{n} = \frac{(\sqrt{n + 16} - \sqrt{n + 9}) (\sqrt{n + 16} + \sqrt{n_{9}})}{\sqrt{n + 16} + \sqrt{n + 9}} = \frac{7}{\sqrt{n + 16} + \sqrt{n + 9}} \\ \underset{\to 0}{\underset{⏟}{\frac{7}{2 \sqrt{2 n}}}} \underset{n > 16}{\leq} c_{n} \leq \underset{\to 0}{\underset{⏟}{\frac{7}{2 \sqrt{n}}}} \\ ⟹ c_{n} \to 0 \end{matrix}

3d

\begin{matrix} d_{n} = \frac{\log (n^{2} + 25)}{\sqrt{n - 4}} \\ Find: lim_{n \to \infty} d_{n} \\ Solution: \\ 0 \leq d_{n} \underset{n > 5}{\leq} \frac{\log (2 n^{2})}{\sqrt{n - 4}} = \frac{\log (2 n^{2})}{\sqrt{n}} = \frac{\log (2) + 2 \log (n)}{\sqrt{n}} = \underset{\to 0}{\underset{⏟}{\frac{\log (2)}{\sqrt{n}}}} + \underset{\to 0}{\underset{⏟}{\frac{2}{\ln (10)} \cdot \underset{\to 0}{\underset{⏟}{\frac{\ln n}{\sqrt{n}}}}}} \\ ⟹ d_{n} \to 0 \end{matrix}

3e

\begin{matrix} p_{n} = \frac{(\log (n + 4))^{81}}{n^{16} + n - 1} \\ Find: lim_{n \to \infty} p_{n} \\ Solution: \\ 0 \leq p_{n} \underset{n - 1 \geq 0}{\leq} \frac{(\log (n + 4))^{81}}{n^{16}} \underset{n > 3}{\leq} \frac{(\log (n^{2}))^{81}}{n^{16}} = {(\frac{2}{\ln (10)})}^{81} \cdot \frac{(\ln n)^{81}}{n^{16}} = \\ = \underset{\to 0}{\underset{⏟}{{(\frac{2}{\ln (10)})}^{81} \cdot \underset{\to 0}{\underset{⏟}{{(\underset{\to 0}{\underset{⏟}{\frac{\ln n}{n^{16 / 81}}}})}^{81}}}}} \\ ⟹ p_{n} \to 0 \end{matrix}

4

\begin{matrix} a_{n} \to 0, b_{n} \to 0 \end{matrix}

4a

\begin{matrix} Find a_{n}, b_{n} : \frac{a_{n}}{b_{n}} \to \infty \\ Solution: \\ a_{n} = \frac{1}{n} \\ b_{n} = \frac{1}{n^{2}} \\ \frac{a_{n}}{b_{n}} = \frac{n^{2}}{n} = n \to \infty \end{matrix}

4b

\begin{matrix} Find a_{n}, b_{n} : \frac{a_{n}}{b_{n}} \to 15 \\ Solution: \\ a_{n} = \frac{15}{n} \\ b_{n} = \frac{1}{n} \\ \frac{a_{n}}{b_{n}} = \frac{15 n}{n} = 15 \to 15 \end{matrix}

4c

\begin{matrix} Find a_{n}, b_{n} : \frac{a_{n}}{b_{n}} \to 0 \\ Solution: \\ a_{n} = \frac{1}{n^{2}} \\ b_{n} = \frac{1}{n} \\ \frac{a_{n}}{b_{n}} = \frac{n}{n^{2}} = \frac{1}{n} \to 0 \end{matrix}

4d

\begin{matrix} Find a_{n}, b_{n} : \frac{a_{n}}{b_{n}} \to - \infty \\ Solution: \\ a_{n} = - \frac{1}{n} \\ b_{n} = \frac{1}{n^{2}} \\ \frac{a_{n}}{b_{n}} = - \frac{n^{2}}{n} = - n \to - \infty \end{matrix}

4e

\begin{matrix} Find a_{n}, b_{n} : ∄ lim_{n \to \infty} \frac{a_{n}}{b_{n}} \\ Solution: \\ a_{n} = \frac{(- 1)^{n}}{n} \\ b_{n} = \frac{1}{n} \\ \frac{a_{n}}{b_{n}} = (- 1)^{n} ⟹ ∄ lim_{n \to \infty} \frac{a_{n}}{b_{n}} \end{matrix}

5a

\begin{matrix} C > 0 \\ Prove by definition: a_{n} \to \infty ⟹ a_{n}^{C} \to \infty \\ Proof: \\ a_{n} \to \infty ⟹ \forall M > 0 : \exists N : \forall n > N : a_{n} > M ⟹ a_{n}^{C} > M^{C} \\ Let M_{1} = M^{C} \\ ⟹ \forall M_{1} > 0 : \exists N : \forall n > N : a_{n}^{C} > M_{1} \\ ⟹ lim_{n \to \infty} a_{n}^{C} = \infty \end{matrix}

5b

\begin{matrix} Give an example: a_{n} \to \infty, c_{n} > 0, a_{n}^{c_{n}} ↛ \infty \\ Solution: \\ a_{n} = e^{n} \\ c_{n} = \frac{1}{n} \\ a_{n}^{c_{n}} = (e^{n})^{1 / n} = e^{n / n} = e^{1} = e \to e \\ ⟹ a_{n}^{c_{n}} ↛ \infty \end{matrix}

6a

\begin{matrix} a_{n} \to 0, b_{n} \to \infty \\ Prove: \frac{a_{n}}{b_{n}} \to 0 \\ Proof: \\ b_{n} \to \infty ⟹ \forall M > 0 : \exists N : \forall n > N : b_{n} > M ⟹ 0 < \frac{1}{b_{n}} < \frac{1}{M} \\ Let ε = \frac{1}{M} \\ \forall ε > 0 : \exists N : \forall n > N : 0 < \frac{1}{b_{n}} < ε = \frac{1}{M} ⟹ \frac{1}{b_{n}} \to 0 \\ lim_{n \to \infty} \frac{a_{n}}{b_{n}} = lim_{n \to \infty} \underset{\to 0}{\underset{⏟}{a_{n}}} \cdot \underset{\to 0}{\underset{⏟}{\frac{1}{b_{n}}}} = 0 \end{matrix}

6b

\begin{matrix} a_{n} \to \infty, b_{n} > 0, b_{n} \to 0 \\ Prove: \frac{a_{n}}{b_{n}} \to \infty \\ Proof: \\ b_{n} > 0, b_{n} \to 0 ⟹ \forall ε > 0 : \exists N : \forall n > N : b_{n} < ε ⟹ \frac{1}{b_{n}} > \frac{1}{ε} \\ Let M = \frac{1}{ε} \\ \forall M > 0 : \exists N : \forall n > N : \frac{1}{b_{n}} > \frac{1}{ε} = M ⟹ \frac{1}{b_{n}} \to \infty \\ lim_{n \to \infty} \frac{a_{n}}{b_{n}} = lim_{n \to \infty} \underset{\to \infty}{\underset{⏟}{a_{n}}} \cdot \underset{\to \infty}{\underset{⏟}{\frac{1}{b_{n}}}} = \infty \end{matrix}

6c

\begin{matrix} a_{n} \to 0, b_{n} \to \infty \\ Prove: a_{n}^{b_{n}} \to 0 \\ Proof: \\ a_{n} \to 0 ⟹ \forall ε > 0 : \exists N_{1} : \forall n > N_{1} : | a_{n} | < ε \\ Let ε = 1 \\ b_{n} \to \infty ⟹ \forall M > 0 : \exists N_{2} : \forall n > N_{2} : b_{n} > M \\ Let M = 1 \\ Let N = m a x (N_{1}, N_{2}) \\ \forall n > N : a_{n} < 1, b_{n} > 1 ⟹ | a_{n} |^{b_{n}} \leq | a_{n} | \\ ⟹ 0 \leq | a_{n}^{b_{n}} | = | a_{n} |^{b_{n}} \leq \underset{\to 0}{\underset{⏟}{| a_{n} |}} \\ ⟹ | a_{n}^{b_{n}} | \to 0 ⟹ a_{n}^{b_{n}} \to 0 \end{matrix}