Exam 2022B (B)

1

\begin{matrix} Let 0 < x < 1 \\ Prove: \frac{\ln (x^{2} + 1)}{\tan x} < \frac{2 x}{x^{2} + 1} \\ Proof: \\ Let f (x) = \ln (x^{2} + 1), f (0) = 0 \\ f is continuous and differentiable on (0, \infty) \\ f^{'} (x) = \frac{2 x}{x^{2} + 1} \\ Let g (x) = \tan x, g (0) = 0 \\ g is continuous and differentiable on (0, \frac{π}{2}) \\ g^{'} (x) = \frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x} = \frac{1}{\cos^{2} x} = 1 + \tan^{2} x \\ By Cauchy’s theorem: \exists c \in (0, x) : \frac{f (x) - f (0)}{g (x) - g (0)} = \frac{f^{'} (c)}{g^{'} (c)} \\ \frac{f^{'} (c)}{g^{'} (c)} = \frac{2 c \cos^{2} c}{c^{2} + 1} \leq \frac{2 c}{c^{2} + 1} \\ Let h (x) = \frac{2 x}{x^{2} + 1} \\ h is continuous and differentiable on R \\ h^{'} (x) = \frac{2 (x^{2} + 1) - 4 x^{2}}{(x^{2} + 1)^{2}} = \frac{2 (1 - x^{2})}{(x^{2} + 1)^{2}} \\ 0 < x < 1 ⟹ (1 - x^{2}) > 0 ⟹ h^{'} (x) > 0 \\ ⟹ [c < x ⟹ h (c) < h (x)] \\ ⟹ \frac{\ln (x^{2} + 1)}{\tan x} = \frac{f (x)}{g (x)} = \frac{f^{'} (c)}{g^{'} (c)} \leq h (c) < h (x) = \frac{2 x}{x^{2} + 1} \end{matrix}

\begin{matrix} Let a_{n} be monotonically non-increasing \\ Prove or disprove: \sum_{n = 1}^{\infty} a_{n} converges ⟹ n a_{n} \to 0 \\ Proof: \\ Let \sum_{n = 1}^{\infty} a_{n} converges \\ ⟹ \sum_{n = 1}^{\infty} 2^{n} a_{2^{n}} converges \\ ⟹ a_{n} \to 0 and 2^{n} a_{2^{n}} \to 0 \\ \forall n \in N : \exists k \in N : 2^{k} \leq n \leq 2^{k + 1} \\ a_{n} is monotonically non-increasing ⟹ a_{2^{k + 1}} \leq a_{n} \leq a_{2^{k}} \\ ⟹ 2^{k} a_{2^{k + 1}} \leq n a_{n} \leq 2^{k + 1} a_{2^{k}} \\ ⟹ \frac{1}{2} \underset{\to 0}{\underset{⏟}{2^{k + 1} a_{2^{k + 1}}}} \leq n a_{n} \leq 2 \cdot \underset{\to 0}{\underset{⏟}{2^{k} a_{2^{k}}}} \\ ⟹ n a_{n} \to 0 \end{matrix}

\begin{matrix} Let a_{n} be monotonically non-increasing \\ Prove or disprove: n a_{n} \to 0 ⟹ \sum_{n = 1}^{\infty} a_{n} converges \\ Disproof: \\ Let a_{n} = \frac{1}{n \ln n} \\ n a_{n} = \frac{1}{\ln n} \to 0 \\ \sum_{n = 1}^{\infty} 2^{n} a_{2^{n}} = \sum_{n = 1}^{\infty} \frac{2^{n}}{2^{n} n \ln (2)} = \frac{1}{\ln (2)} \sum_{n = 1}^{\infty} \frac{1}{n} diverges \\ ⟹ \sum_{n = 1}^{\infty} a_{n} diverges \end{matrix}

\begin{matrix} \sum_{n = 1}^{\infty} \frac{\sqrt[n]{7^{n} + 19^{n}}}{n^{3} \sin (\frac{1}{n})} \\ Solution: \\ Let a_{n} = \frac{\sqrt[n]{7^{n} + 19^{n}}}{n^{3} \sin (\frac{1}{n})} \\ \forall n \in N : 0 < \frac{1}{n} \leq 1 \\ \forall x \in (0, 1] : \sin (x) > 0 ⟹ \sin (\frac{1}{n}) > 0 ⟹ | \sin (\frac{1}{n}) | = \sin (\frac{1}{n}) \\ Let b_{n} = \frac{1}{n^{2}} \\ lim_{n \to \infty} | \frac{a_{n}}{b_{n}} | = \frac{\sqrt[n]{7^{n} + 19^{n}}}{n \sin (\frac{1}{n})} = lim_{n \to \infty} \frac{\sqrt[n]{7^{n} + 19^{n}}}{\underset{\to 1}{\underset{⏟}{\frac{\sin (\frac{1}{n})}{\frac{1}{n}}}}} = lim_{n \to \infty} \sqrt[n]{7^{n} + 19^{n}} \\ Let c_{n} = 7^{n} + 19^{n} \\ lim_{n \to \infty} \frac{c_{n + 1}}{c_{n}} = lim_{n \to \infty} \frac{7^{n + 1} + 19^{n + 1}}{7^{n} + 19^{n}} = lim_{n \to \infty} \frac{7 \cdot \overset{\to 0}{\overset{⏞}{{(\frac{7}{19})}^{n}}} + 19}{\underset{\to 0}{\underset{⏟}{{(\frac{7}{19})}^{n}}} + 1} = \frac{19}{1} = 19 \\ ⟹ lim_{n \to \infty} \sqrt[n]{7^{n} + 19^{n}} = 19 \\ ⟹ lim_{n \to \infty} | \frac{a_{n}}{b_{n}} | = 19 \\ ⟹ By the limit comparison test: [\sum_{n = 1}^{\infty} b_{n} converges ⟺ \sum_{n = 1}^{\infty} a_{n} converges] \\ \sum_{n = 1}^{\infty} b_{n} converges ⟹ \sum_{n = 1}^{\infty} a_{n} also converges \\ ⟹ \sum_{n = 1}^{\infty} \frac{\sqrt[n]{7^{n} + 19^{n}}}{n^{3} \sin (\frac{1}{n})} converges absolutely \end{matrix}

\begin{matrix} \sum_{n = 1}^{\infty} \frac{n + n^{2} + n^{3} + \dots + n^{n}}{n^{n + 1}} \\ Solution: \\ \frac{n + n^{2} + \dots + n^{n}}{n^{n + 1}} = \frac{1}{n^{n}} + \frac{1}{n^{n - 1}} + \dots + \frac{1}{n} \geq \frac{1}{n} \\ \sum_{n = 1}^{\infty} \frac{1}{n} diverges \\ ⟹ By the direct comparison test: \sum_{n = 1}^{\infty} \frac{n + n^{2} + n^{3} + \dots + n^{n}}{n^{n + 1}} diverges \end{matrix}

\begin{matrix} Let f be a function \\ Let f be continuous on [a, b] \\ Let f be differentiable on (a, b) \\ Prove or disprove: f^{'} is bounded on (a, b) \\ Disproof: \\ f^{'} doesn’t have to be continuous \\ Let f (x) = {\begin{cases} x^{5} \sin (\frac{1}{x^{7}}) & x \neq 0 \\ 0 & x = 0 \end{cases} \\ f is continuous on [- 1, 1] \\ f^{'} (0) = lim_{x \to 0} \frac{f (x) - f (0)}{x} = lim_{x \to 0} x^{4} \sin (\frac{1}{x^{7}}) = 0 \\ ⟹ f^{'} (x) = {\begin{cases} 5 x^{4} \sin (\frac{1}{x^{7}}) - \frac{3}{x^{3}} \cos (\frac{1}{x^{7}}) & x \neq 0 \\ 0 & x = 0 \end{cases} \\ x \to 0 ⟹ \cos (\frac{1}{x^{7}}) ↛ 0 ⟹ \frac{3}{x^{3}} \cos (\frac{1}{x^{7}}) \to \infty \\ ⟹ f^{'} (x) is not bounded on [- 1, 1] \end{matrix}

\begin{matrix} Let f be a function \\ Let f be continuous on [0, 1] \\ Let f (0) = f (1) \\ Let n \in N \\ Prove or disprove: \exists c \in [0, 1] : f (c) = f (c + \frac{1}{n}) \\ Proof: \\ Let g (x) = f (x) - f (x + \frac{1}{n}) \\ g (0) + g (\frac{1}{n}) + g (\frac{2}{n}) + \dots + g (\frac{n - 1}{n}) = \\ = f (0) - f (\frac{1}{n}) + f (\frac{1}{n}) - f (\frac{2}{n}) + \dots + f (\frac{n - 1}{n}) - f (1) = f (0) - f (1) = 0 \\ If any of additives is 0 \exists c \in [0, 1] : g (c) = 0 and we are done \\ If all additives are not 0, then there must be at least one negative and one positive \\ ⟹ By the intermediate value theorem exists c \in [a, b] : g (c) = 0 and we are done \end{matrix}

\begin{matrix} Let a_{n} = {\begin{cases} 7 & n = 1 \\ \sin (a_{n - 1}) & n > 1 \end{cases} \\ Find lim_{n \to \infty} a_{n} \\ Solution: \\ a_{2} = \sin (a_{1}) = \sin (7) \\ 0 < \sin (7) \leq 1 \\ a_{3} = \sin (\sin (a_{2})) \\ 0 < x \leq 1 ⟹ \sin (x) > 0 \\ \sin (x) < x (Geometrical proof from lectures) \\ ⟹ a_{n + 1} < a_{n} ⟹ a_{n} is monotonically decreasing and bounded from below by 0 \\ ⟹ a_{n} converges \\ Let lim_{n \to \infty} a_{n} = L \\ lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} \sin (a_{n}) = lim_{n \to \infty} \sin (L) = \sin (L) \\ ⟹ L = \sin (L) \\ \forall n > 1 : 0 < a_{n} \leq 1 ⟹ 0 \leq L \leq 1 \\ L > 0 ⟹ \sin (L) < L ⟹ L = 0 ⟹ a_{n} \to 0 \end{matrix}