Infi-1 6

\begin{matrix} a_{n} > 0 \\ lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = L ⟹ lim_{n \to \infty} \sqrt[n]{a_{n}} = L \end{matrix}

Exercise

\begin{matrix} lim_{n \to \infty} \sqrt[n]{n} = ? \\ a_{n} = n \\ lim_{n \to \infty} \sqrt[n]{n} \underset{\sqrt[n]{n} > 0}{=} lim_{n \to \infty} \frac{n + 1}{n} = lim_{n \to \infty} 1 + \underset{\to 0}{\underset{⏟}{\frac{1}{n}}} = 1 \end{matrix}

\begin{matrix} Prove or disprove: lim_{n \to \infty} a_{n} = \infty ⟹ a_{n} is monotonically non-descending \\ Disproof: \\ Let a_{n} = n + (- 1)^{n} \\ a_{n + 1} - a_{n} = 1 + (- 1)^{n + 1} - (- 1)^{n} = 1 - 2 (- 1)^{n} ≮ 0 \\ lim_{n \to \infty} a_{n} = \infty ⟹ Disproved \end{matrix}

\begin{matrix} Prove or disprove: lim_{n \to \infty} a_{n} = 0 ⟹ lim_{n \to \infty} \frac{1}{a_{n}} = \infty \\ Proved in seminar Infi-1 5 \end{matrix}

\begin{matrix} Prove or disprove: lim_{n \to \infty} a_{n} b_{n} = 0 ⟹ a_{n} is bounded or b_{n} is bounded \\ Disproof: \\ a_{n} = {\begin{cases} 0 & n = 2 k \\ n & otherwise \end{cases}, b_{n} = {\begin{cases} 0 & n = 2 k + 1 \\ n & otherwise \end{cases} \\ a_{n} b_{n} = 0 \end{matrix}

\begin{matrix} x > 0 \\ a_{n} = \frac{6 n + \sqrt{⌊ x^{2} n^{2} ⌋}}{3 n + \sqrt{2}} \\ Prove or disprove: lim_{n \to \infty} a_{n} > 2 \\ Proof: \\ \frac{6 n + \sqrt{x^{2} n^{2} - 1}}{3 n + \sqrt{2}} < \frac{6 n + \sqrt{⌊ x^{2} n^{2} ⌋}}{3 n + \sqrt{2}} < \frac{6 n + \sqrt{x^{2} n^{2}}}{3 n + \sqrt{2}} \\ 2 + \frac{x}{3} \leftarrow \frac{6 + \overset{\to x}{\overset{⏞}{\sqrt{x^{2} - \overset{\to 0}{\overset{⏞}{\frac{1}{n^{2}}}}}}}}{3 + \underset{\to 0}{\underset{⏟}{\frac{\sqrt{2}}{n}}}} < a_{n} < \frac{6 + x}{3 + \underset{\to 0}{\underset{⏟}{\frac{\sqrt{2}}{n}}}} \to 2 + \frac{x}{3} \\ ⟹ lim_{n \to \infty} a_{n} = 2 + \underset{> 0}{\underset{⏟}{\frac{x}{3}}} ⟹ lim_{n \to \infty} a_{n} > 2 \end{matrix}

\begin{matrix} Prove or disprove: \\ 1. & \exists a_{n} : lim_{n \to \infty} a_{n} = \infty \land lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = 0 \\ 2. & \exists a_{n} : lim_{n \to \infty} a_{n} = \infty \land lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = 4 \\ 3. & \exists a_{n} : lim_{n \to \infty} a_{n} = \infty \land lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = \infty \\ 4. & \exists a_{n} : lim_{n \to \infty} a_{n} = \infty \land ∄ lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} \end{matrix}

\begin{matrix} 1. Disproof: \\ Let lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = 0 \\ ⟹ \exists N : \forall n > N : \frac{a_{n + 1}}{a_{n}} < 1 ⟹ a_{n} is monotonically descending after N \\ ⟹ lim_{n \to \infty} a_{n} \neq \infty \end{matrix}

\begin{matrix} 2. Proof: \\ a_{n} = 4^{n} ⟹ \frac{a_{n + 1}}{a_{n}} = 4 \to 4 \end{matrix}

\begin{matrix} 3. Proof: \\ a_{n} = n! ⟹ \frac{a_{n + 1}}{a_{n}} = n + 1 \to \infty \end{matrix}

\begin{matrix} 4. Proof: \\ a_{n} = {\begin{cases} n & n = 2 k \\ n^{2} & otherwise \end{cases} \\ a_{n} \geq n \to \infty ⟹ a_{n} \to \infty \\ \frac{a_{n + 1}}{a_{n}} = {\begin{cases} \frac{1}{n} + \frac{1}{n^{2}} & n = 2 k \\ n + 2 + \frac{1}{n} & otherwise \end{cases} ⟹ ∄ lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} \end{matrix}

\begin{matrix} {\begin{matrix} a_{1} = 5 \\ a_{n + 1} = a_{n} \cdot \frac{6 + a_{n}}{3 + 2 a_{n}} \end{matrix} \\ Base case. a_{1} = 5 \geq 3 \\ Induction step. Let a_{n} \geq 3 \\ a_{n + 1} = a_{n} \cdot \frac{6 + a_{n}}{3 + 2 a_{n}} \geq a_{n} \cdot \frac{6 + 3}{3 + 2 a_{n}} = \frac{9 a_{n}}{3 + 2 a_{n}} \geq \frac{9 a_{n}}{a_{n} + 2 a_{n}} = 3 \\ ⟹ a_{n + 1} \geq 3 \\ By induction: a_{n} \geq 3 \\ a_{n + 1} - a_{n} = a_{n} \frac{3 - a_{n}}{3 + 2 a_{n}} < 0 ⟹ a_{n} is monotonically descending \\ ⟹ a_{n} converges \\ Let lim_{n \to \infty} a_{n} = L \\ L = lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} a_{n} \cdot \frac{6 + a_{n}}{3 + 2 a_{n}} = L \cdot \frac{6 + L}{3 + 2 L} \\ ⟹ [\begin{matrix} L = 0 \\ L = 3 \end{matrix} \\ a_{n} \geq 3 ⟹ lim_{n \to \infty} a_{n} \neq 0 ⟹ lim_{n \to \infty} a_{n} = 3 \end{matrix}