Exam 2024 (A)

1a

\begin{matrix} Prove: x > 3 ⟹ \frac{2 \ln (x - 2)}{\sqrt{x + 1} - 2} > \frac{4 \sqrt{x + 1}}{x - 2} \\ Proof: \\ Let f (t) = 2 \ln (t - 2) \\ f^{'} (t) = \frac{2}{t - 2} \\ Let g (t) = \sqrt{t + 1} - 2 \\ g^{'} (t) = \frac{1}{2 \sqrt{t + 1}} \\ f (t) is continuous on [3, \infty) and differentiable on (3, \infty) \\ g (t) is continuous on [3, \infty] and differentiable on (3, \infty) \\ ⟹ By Lagrange’s theorem: \exists c \in (3, x) : \frac{f (x) - f (3)}{g (x) - g (3)} = \frac{f^{'} (c)}{g^{'} (c)} \\ ⟹ \frac{2 \ln (x - 2) - 2 \ln (1)}{\sqrt{x + 1} - 2 + \sqrt{4} - 2} = \frac{2 \ln (x - 2)}{\sqrt{x + 1} - 2} = \frac{4 \sqrt{c + 1}}{c - 2} \\ Let h (t) = \frac{4 \sqrt{t + 1}}{t - 2} \\ h^{'} (t) = \frac{(\frac{2}{\sqrt{t + 1}} (t - 2) - 4 \sqrt{t + 1})}{(t - 2)^{2}} = \frac{2 t - 4 - 4 t - 4}{\sqrt{t + 1} (t - 2)^{2}} = \frac{\overset{< 0}{\overset{⏞}{- 2 t - 8}}}{\underset{> 0}{\underset{⏟}{\sqrt{t + 1}}} \underset{> 0}{\underset{⏟}{(t - 2)^{2}}}} < 0 \\ ⟹ h (t) is monotonically decreasing for t > 3 \\ c < x ⟹ h (c) > h (x) ⟹ \frac{2 \ln (x - 2)}{\sqrt{x + 1} - 2} > \frac{4 \sqrt{x + 1}}{x - 2} \end{matrix}

\begin{matrix} Let f be a function \\ Let f be continuous on [a, b] and differentiable on (a, b) \\ Prove or disprove: f is bounded on (a, b) \\ Proof: \\ Let f be unbounded on (a, b) \\ ⟹ \forall n \in N : \exists x_{n} \in (a, b) : f (x_{n}) > n \\ (a, b) is bounded ⟹ By Bolzano-Weierstrasse theorem \exists x_{n_{k}} \to x \in (a, b) \\ f is continuous on [a, b] ⟹ f (x_{n_{k}}) \to f (x) \in R \\ f (x_{n}) \to \infty ⟹ f (x_{n_{k}}) \to \infty ⟹ f (x) = \infty - Contradiction! \\ ⟹ f is bounded on (a, b) \\ Note: This is Weierstrasse boundness theorem (lecture 23) \end{matrix}

\begin{matrix} Find: lim_{n \to \infty} {(1 + \cos (\frac{1}{n}) \sin (\frac{1}{n}))}^{n^{2} \sin (1 / n)} \\ Solution: \\ Let a_{n} = \cos (\frac{1}{n}) \sin (\frac{1}{n}) \\ Let b_{n} = n^{2} \sin (\frac{1}{n}) \\ \sin (\frac{1}{n}) \to \sin 0 = 0 \\ - 1 < \cos (\frac{1}{n}) < 1 \\ ⟹ \cos (\frac{1}{n}) \sin (\frac{1}{n}) \to 0 \\ ⟹ a_{n} \to 1 \\ b_{n} \to \infty \\ ⟹ lim_{n \to \infty} a_{n}^{b_{n}} = e^{lim_{n \to \infty} (a_{n} - 1) b_{n}} \\ lim_{n \to \infty} (a_{n} - 1) b_{n} = lim_{n \to \infty} \cos (\frac{1}{n}) \sin (\frac{1}{n}) n^{2} \sin (\frac{1}{n}) = \\ = lim_{n \to \infty} \frac{\sin (\frac{1}{n})}{\frac{1}{n}} \frac{\sin (\frac{1}{n})}{\frac{1}{n}} \cos (\frac{1}{n}) \underset{\frac{1}{n} \to 0^{+}}{\underset{⏟}{=}} 1 \\ ⟹ lim_{n \to \infty} a_{n}^{b_{n}} = e^{1} = e \end{matrix}

\begin{matrix} Find: lim_{n \to \infty} {(1 + \frac{1}{n})}^{(- 1)^{n}} \\ Solution: \\ Let a_{n} = {(1 + \frac{1}{n})}^{(- 1)^{n}} \\ a_{2 n} = (1 + \frac{1}{2 n}) \to 1 \\ a_{2 n - 1} = {(1 + \frac{1}{2 n - 1})}^{- 1} = \frac{2 n - 1}{2 n} = 1 - \frac{1}{2 n} \to 1 \\ {\begin{matrix} a_{2 n} \to 1 \\ a_{2 n - 1} \to 1 \end{matrix} ⟹ a_{n} \to 1 \end{matrix}

\begin{matrix} Determine whether series diverges or converges \\ and if converges, whether absolutely or conditionally \\ \sum_{n = 1}^{\infty} \frac{17}{(\sqrt[n]{n!})^{17}} \\ Solution: \\ Let a_{n} = \frac{17}{(\sqrt[n]{n!})^{17}} \\ | a_{n} | = a_{n} \\ n! = \prod_{i = 1}^{n} i \geq \prod_{i = \frac{n}{2}}^{n} i \geq \prod_{i = \frac{n}{2}}^{n} \frac{n}{2} = {(\frac{n}{2})}^{n / 2} \\ ⟹ \sqrt[n]{n!} \geq \sqrt[n]{{(\frac{n}{2})}^{n / 2}} = \sqrt{(\frac{n}{2})} \\ ⟹ a_{n} \leq \frac{17 \cdot 2^{17 / 2}}{n^{17 / 2}} \\ \sum_{n = 1}^{\infty} \frac{17 \cdot 2^{17 / 2}}{n^{17 / 2}} = 17 \cdot 2^{17 / 2} \cdot \sum_{n = 1}^{\infty} \frac{1}{n^{17 / 2}} converges \\ ⟹ \sum_{n = 1}^{\infty} | a_{n} | converges \\ ⟹ \sum_{n = 1}^{\infty} \frac{17}{(\sqrt[n]{n!})^{17}} converges absolutely \end{matrix}

\begin{matrix} Determine whether series diverges or converges \\ and if converges, whether absolutely or conditionally \\ \sum_{n = 2}^{\infty} \frac{(- 1)^{n}}{\ln (n^{\sqrt{n}})} \\ Solution: \\ Let a_{n} = \frac{(- 1)^{n}}{\ln (n^{\sqrt{n}})} \\ | a_{n} | = \frac{1}{\ln (n^{\sqrt{n}})} = \frac{1}{\sqrt{n} \ln n} \\ \sum_{n = 2}^{\infty} \frac{2^{n}}{\sqrt{2^{n}} \ln (2^{n})} = \sum_{n = 2}^{\infty} \frac{\sqrt{2^{n}}}{n \ln (2)} \\ \frac{\sqrt{2^{n}}}{n} ↛ 0 ⟹ \sum_{n = 2}^{\infty} 2^{n} | a_{2^{n}} | diverges ⟹ \sum_{n = 2}^{\infty} | a_{n} | diverges \\ \frac{1}{\ln (n^{\sqrt{n}})} = \frac{1}{\sqrt{n} \ln (n)} which is monotonically decreasing and \to 0 \\ ⟹ By the Leibniz criterion: \sum_{n = 2}^{\infty} \frac{(- 1)^{n}}{\ln (n^{\sqrt{n}})} converges conditionally \end{matrix}

\begin{matrix} Let a_{n} = {\begin{cases} 3 & n = 1 \\ \frac{1}{2} (a_{n - 1} + \frac{7}{a_{n - 1}}) & n > 1 \end{cases} \\ Prove that a_{n} converges and find lim_{n \to \infty} a_{n} \\ Solution: \\ \forall n \in N : a_{n} > 0 \\ a_{n + 1} - a_{n} = \frac{7 - a_{n}^{2}}{2 a_{n}} \\ Base case. \sqrt{7} < a_{1} \leq 3 \\ Induction step. Let \sqrt{7} < a_{n} \leq 3 \\ ⟹ a_{n + 1} - a_{n} < 0 ⟹ a_{n + 1} < a_{n} \\ a_{n + 1} = \frac{a_{n}^{2} + 7}{2 a_{n}} \\ Let f (x) = \frac{x^{2} + 7}{2 x} \\ f is continuous and differentiable on (0, \infty) \\ f^{'} (x) = \frac{1}{2} - \frac{7}{2 x^{2}} \\ x > \sqrt{7} ⟹ f^{'} (x) > 0 ⟹ f (x) > f (\sqrt{7}) = \sqrt{7} \\ a_{n} > \sqrt{7} ⟹ a_{n + 1} = f (a_{n}) > \sqrt{7} \\ ⟹ \forall n \in N : a_{n} > \sqrt{7} \land a_{n + 1} < a_{n} \\ ⟹ a_{n} is bounded and monotonically decreasing \\ ⟹ lim_{n \to \infty} a_{n} = L \in R \\ ⟹ lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} \frac{a_{n}}{2} + \frac{7}{2 a_{n}} = \frac{L}{2} + \frac{7}{2 L} = L \\ ⟹ L^{2} = 7 ⟹ L = \pm \sqrt{7} \\ a_{n} > \sqrt{7} ⟹ L \geq \sqrt{7} ⟹ L = \sqrt{7} \end{matrix}

\begin{matrix} Let S, T \subseteq R, S \subseteq T \\ Prove or disprove: s u p (S) \leq s u p (T) \\ Disproof: \\ Let S = T = R \\ ∄ s u p (S), s u p (T) \end{matrix}