Exam 2023B (B)

1

\begin{matrix} Let x > 0 \\ Prove: \frac{\arctan (\frac{x}{x + 1})}{\ln (x + 1)} > \frac{x + 1}{2 x^{2} + 2 x + 1} \\ Proof: \\ Let f (x) = \arctan (\frac{x}{x + 1}) \\ f^{'} (x) = {(\frac{x}{x + 1})}^{'} \cdot \frac{1}{{(\frac{x}{x + 1})}^{2} + 1} = \frac{x + 1 - x}{(x + 1)^{2}} \cdot \frac{1}{\frac{2 x^{2} + 2 x + 1}{x^{2} + 2 x + 1}} = \frac{1}{2 x^{2} + 2 x + 1} \\ Let g (x) = \ln (x + 1) \\ g^{'} (x) = \frac{1}{x + 1} \\ \frac{f^{'} (x)}{g^{'} (x)} = \frac{x + 1}{2 x^{2} + 2 x + 1} \\ \frac{f (x)}{g (x)} = \frac{f (x) - 0}{g (x) - 0} = \frac{f (x) - f (0)}{g (x) - g (0)} \\ By Langranges theorem: \exists c \in (0, x) : \frac{f (x) - f (0)}{g (x) - g (0)} = \frac{f^{'} (c)}{g^{'} (c)} \\ Let h (x) = \frac{f^{'} (x)}{g^{'} (x)} \\ h^{'} (x) = \frac{2 x^{2} + 2 x + 1 - (x + 1) (4 x + 2)}{(2 x^{2} + 2 x + 1)^{2}} = \frac{- 2 x^{2} - 4 x - 1}{(2 x^{2} + 2 x + 1)^{2}} < 0 \\ ⟹ [c < x ⟹ h (c) > h (x)] \\ ⟹ \frac{f (x)}{g (x)} > h (x) ⟹ \frac{\arctan (\frac{x}{x + 1})}{\ln (x + 1)} > \frac{x + 1}{2 x^{2} + 2 x + 1} \end{matrix}

\begin{matrix} Let f be a function defined on (a, b) \\ Let f be differentiable three times on (a, b) \\ Let \exists c \in (a, b) : f^{'} (c) = 0 \\ Prove or disprove: f^{″} (c) > 0 ⟹ c is a local minimum \\ Proof: \\ f^{″} (c) = lim_{h \to 0} \frac{f^{'} (c + h) - f^{'} (c)}{h} = lim_{h \to 0} \frac{f^{'} (c + h)}{h} > 0 \\ h \to 0^{+} ⟹ f^{'} (c + h) > 0 ⟹ \exists ε_{R} > 0 : f is monotonically increasing on [c, c + ε_{E}] \\ h \to 0^{-} ⟹ f^{'} (c + h) < 0 ⟹ \exists ε_{L} > 0 : f is monotonically decreasing on [c - ε_{L}, c] \\ ⟹ \exists ε = m i n (ε_{L}, ε_{R}) > 0 : \forall x \in [c - ε, c + ε] : f (c) \leq f (x) \\ ⟹ c is a local minimum of f \end{matrix}

\begin{matrix} Let f be a function defined on (a, b) \\ Let f be differentiable three times on (a, b) \\ Let \exists c \in (a, b) : f^{'} (c) = 0 \\ Prove or disprove: c is a local maximum ⟹ f^{″} (c) < 0 \\ Disproof: \\ f (x) = - x^{4} \\ f^{'} (x) = - 4 x^{3} \\ f^{″} (x) = - 12 x^{2} \\ 0 is a local maximum of f \\ f^{″} (0) = 0 \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 a_{1}, b_{1} > 0 \in R \\ Let a_{n + 1} = \frac{a_{n} + b_{n}}{2} \\ Let b_{n + 1} = \sqrt{a_{n} b_{n}} \\ Prove: lim_{n \to \infty} a_{n} = lim_{n \to \infty} b_{n} = L \in R \\ Proof: \\ Base case. a_{1} > 0, b_{1} > 0 \\ Induction step. Let a_{n} > 0, b_{n} > 0 \\ a_{n + 1} = \frac{a_{n} + b_{n}}{2} > 0 \\ b_{n + 1} = \sqrt{a_{n} b_{n}} > 0 \\ ⟹ \forall n \in N : a_{n} > 0, b_{n} > 0 \\ \frac{a_{n} + b_{n}}{2} - \sqrt{a_{n} b_{n}} = \frac{a_{n} - 2 \sqrt{a_{n} b_{n}} + b_{n}}{2} = \frac{(\sqrt{a_{n}} - \sqrt{b_{n}})^{2}}{2} > 0 \\ ⟹ a_{n + 1} > b_{n + 1} \\ ⟹ \forall n > 1 \in N : b_{n} < a_{n} \\ ⟹ \forall n > 1 \in N : a_{n + 1} < \frac{2 a_{n}}{2} = a_{n} \\ ⟹ a_{n} is monotonically decreasing and bounded from below by 0 \\ ⟹ \exists lim_{n \to \infty} a_{n} = L \in R \\ b_{n} = 2 a_{n + 1} - a_{n} ⟹ lim_{n \to \infty} b_{n} = lim_{n \to \infty} 2 a_{n + 1} - a_{n} = 2 L - L = L \\ ⟹ lim_{n \to \infty} a_{n} = lim_{n \to \infty} b_{n} = L \in R \end{matrix}

\begin{matrix} \sum_{n = 1}^{\infty} \frac{\sin n}{n} \\ Solution: \\ Let S_{N} = \sum_{n = 1}^{N} \sin n \\ 2 \sin (\frac{1}{2}) S_{N} = \sum_{n = 1}^{N} 2 \sin (\frac{1}{2}) \sin n = \sum_{n = 1}^{N} (\cos (n - \frac{1}{2}) - \cos (n + \frac{1}{2})) = \\ = \cos (\frac{1}{2}) - \cos (N + \frac{1}{2}) \\ ⟹ S_{N} = \frac{\cos (\frac{1}{2}) - \cos (N + \frac{1}{2})}{2 \sin (\frac{1}{2})} \\ ⟹ \frac{\cos (\frac{1}{2}) - 1}{2 \sin (\frac{1}{2})} \leq S_{N} \leq \frac{\cos (\frac{1}{2})}{2 \sin (\frac{1}{2})} \\ ⟹ \sum_{n = 1}^{\infty} \sin (n) is bounded \\ \frac{1}{n} is monotonically decreasing and \frac{1}{n} \to 0 \\ ⟹ By Dirichlet’s test: \sum_{n = 1}^{\infty} \frac{\sin n}{n} converges \\ \sum_{n = 1}^{\infty} | \frac{\sin n}{n} | = \sum_{n = 1}^{\infty} \frac{| \sin n |}{n} \geq \sum_{n = 1}^{\infty} \frac{1}{2 n} + X \\ \frac{1}{2} \sum_{n = 1}^{\infty} \frac{1}{n} diverges \\ ⟹ \sum_{n = 1}^{\infty} \frac{| \sin n |}{n} also diverges \\ ⟹ \sum_{n = 1}^{\infty} \frac{\sin n}{n} converges conditionally \end{matrix}

\begin{matrix} \sum_{n = 2}^{\infty} \frac{(- 1)^{n}}{\ln (n!)} \\ Solution: \\ \frac{1}{\ln (n!)} is monotonically decreasing and \frac{1}{\ln (n!)} \to 0 \\ ⟹ \sum_{n = 2}^{\infty} \frac{(- 1)^{n}}{\ln (n!)} converges at least conditionally by the alternating series test \\ \sum_{n = 2}^{\infty} | \frac{(- 1)^{n}}{\ln (n!)} | = \sum_{n = 2}^{\infty} \frac{1}{\ln (n!)} \\ k < n ⟹ \ln (k) < \ln (n) ⟹ \ln (n!) = \ln (1) + \ln (2) + \dots + \ln (n) \leq n \ln n \\ ⟹ \frac{1}{\ln (n!)} \geq \frac{1}{n \ln n} \\ \sum_{n = 2}^{\infty} \frac{1}{n \ln n} diverges ⟹ \sum_{n = 2}^{\infty} \frac{1}{\ln (n!)} also diverges \\ ⟹ \sum_{n = 2}^{\infty} \frac{(- 1)^{n}}{\ln (n!)} coverges conditionally \end{matrix}

\begin{matrix} \sum_{n = 1}^{\infty} \frac{1}{(n + 2) (n + 4)} \\ Solution: \\ \frac{1}{(n + 2) (n + 4)} = \frac{1}{2} (\frac{n + 4}{(n + 2) (n + 4)} - \frac{n + 2}{(n + 2) (n + 4)}) = \frac{1}{2} (\frac{1}{n + 2} - \frac{1}{n + 4}) \\ ⟹ \sum_{n = 1}^{\infty} \frac{1}{(n + 2) (n + 4)} = \frac{1}{2} \sum_{n = 1}^{\infty} (\frac{1}{n + 2} - \frac{1}{n + 4}) \\ ⟹ 2 S_{N} = \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6} + \frac{1}{5} - \frac{1}{7} + \dots + \frac{1}{N + 1} - \frac{1}{N + 3} + \frac{1}{N + 2} - \frac{1}{N + 4} = \\ = \frac{1}{3} + \frac{1}{4} - \frac{1}{N + 3} - \frac{1}{N + 4} \\ ⟹ S_{N} = \frac{1}{6} + \frac{1}{8} - \frac{1}{2 N + 6} - \frac{1}{2 N + 8} \to \frac{1}{6} + \frac{1}{8} \\ ⟹ \sum_{n = 1}^{\infty} \frac{1}{(n + 2) (n + 4)} = \frac{1}{6} + \frac{1}{8} = \frac{14}{48} \end{matrix}

\begin{matrix} lim_{n \to \infty} (e^{1 / n} - e^{- 1 / n})^{\sin (1 / n)} \\ Solution: \\ Let t = \frac{1}{n} \\ lim_{t \to 0} (e^{t} - e^{- t})^{\sin t} = lim_{t \to 0} \frac{(e^{2 t} - 1)^{\sin t}}{e^{t \sin t}} = lim_{t \to 0} (e^{2 t} - 1)^{\sin t} = lim_{t \to 0} e^{\sin t \ln (e^{2 t} - 1)} \\ lim_{t \to 0} \sin t \ln (e^{2 t} - 1) = lim_{t \to 0} t \ln (e^{2 t} - 1) \cdot \frac{\sin t}{t} = lim_{t \to 0} \frac{\ln (e^{2 t} - 1)}{\frac{1}{t}} \overset{L}{=} lim_{t \to 0} \frac{\frac{2 e^{2 t}}{e^{2 t} - 1}}{- \frac{1}{t^{2}}} = \\ = - lim_{t \to 0} \frac{2 e^{2 t} t^{2}}{e^{2 t} - 1} \overset{L}{=} - 2 lim_{t \to 0} \frac{\overset{\to 0}{\overset{⏞}{2 e^{2 t} t^{2}}} + \overset{\to 0}{\overset{⏞}{2 e^{2 t} t}}}{\underset{\to 1}{\underset{⏟}{2 e^{2 t}}}} = - 2 \cdot 0 = 0 \\ ⟹ lim_{n \to \infty} (e^{1 / n} - e^{- 1 / n})^{\sin (1 / n)} = 0 \end{matrix}