Exam 2022B (A)

1

\begin{matrix} Let 0 < x < \sqrt{3} \\ Prove: \frac{\sqrt{x}}{\ln (x^{2} + 1)} > \frac{x^{2} + 1}{4 x \sqrt{x}} \\ Proof: \\ Let f (x) = \sqrt{x} \\ f is continuous and differentiable on (0, \infty) \\ f^{'} (x) = \frac{1}{2 \sqrt{x}} \\ Let g (x) = \ln (x^{2} + 1) \\ g is continuous and differentiable on (0, \infty) \\ g^{'} (x) = \frac{2 x}{x^{2} + 1} \\ \frac{f (x)}{g (x)} = \frac{f (x) - 0}{g (x) - 0} = \frac{f (x) - f (0)}{g (x) - g (0)} \\ By the Langange’s 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)} = \frac{x^{2} + 1}{4 x \sqrt{x}} \\ h is continuous and differentiable on (0, \infty) \\ h^{'} (x) = \frac{8 x^{2} \sqrt{x} - (x^{2} + 1) 6 \sqrt{x}}{16 x^{3}} = \frac{8 x^{2} \sqrt{x} - 6 x^{2} \sqrt{x} - 6 \sqrt{x}}{16 x^{3}} = \frac{x^{2} \sqrt{x} - 3 \sqrt{x}}{8 x^{3}} = \\ = \frac{x^{2} - 3}{8 \sqrt{x^{5}}} \\ 0 < x < \sqrt{3} ⟹ h (x) < 0 ⟹ [c < x ⟹ h (c) > h (x)] \\ ⟹ \frac{f (x)}{g (x)} = \frac{f^{'} (c)}{g^{'} (c)} = h (c) > h (x) = \frac{f^{'} (x)}{g^{'} (x)} \\ ⟹ \frac{\sqrt{x}}{\ln (x^{2} + 1)} > \frac{x^{2} + 1}{4 x \sqrt{x}} \end{matrix}

2a

\begin{matrix} Prove or disprove: a_{n} = \sum_{k = n}^{2 n} \frac{1}{k} converges \\ Proof: \\ a_{n + 1} - a_{n} = \sum_{k = n}^{2 n} \frac{1}{k} - \sum_{k = n + 1}^{2 n + 2} \frac{1}{k} = \frac{1}{2 n + 1} + \frac{1}{2 n + 2} - \frac{1}{n} \leq \frac{2}{2 n} - \frac{1}{n} = 0 \\ ⟹ a_{n + 1} \leq a_{n} ⟹ a_{n} is monotonically decreasing \\ a_{n} \geq \sum_{k = n}^{2 n} \frac{1}{n} = 1 ⟹ a_{n} is lower bounded \\ ⟹ a_{n} converges \end{matrix}

2b

\begin{matrix} Let a_{n} > 0 \\ Let lim_{n \to \infty} \sqrt[n]{a_{n}} = L > 0 \\ Prove or disprove: lim_{n \to \infty} \frac{a_{n}}{a_{n + 1}} = \frac{1}{L} \\ Disproof: \\ Let a_{n} = 1, 5, 1, 5, \dots \\ lim_{n \to \infty} \sqrt[2 n]{a_{2 n}} = lim_{n \to \infty} \sqrt[2 n]{5} = 1 \\ lim_{n \to \infty} \sqrt[2 n - 1]{a_{2 n - 1}} = lim_{n \to \infty} \sqrt[2 n - 1]{1} = 1 \\ ⟹ \sqrt[n]{a_{n}} \to 1 \\ \frac{a_{n}}{a_{n + 1}} = \frac{1}{5}, 5, \frac{1}{5}, 5, \dots ⟹ f \end{matrix}

3a

\begin{matrix} \sum_{n = 1}^{\infty} \frac{(- 1)^{n} n}{n^{\sqrt{n}}} \\ Solution: \\ Let a_{n} = \frac{(- 1)^{n} n}{n^{\sqrt{n}}} \\ \sum_{n = 1}^{\infty} | a_{n} | = \sum_{n = 1}^{\infty} \frac{n}{n^{\sqrt{n}}} = \sum_{n = 1}^{\infty} \frac{1}{n^{\sqrt{n} - 1}} \\ \forall n > 9 : \sqrt{n} - 1 > 2 ⟹ \sum_{n = 1}^{\infty} = \sum_{n = 1}^{9} \frac{1}{n^{\sqrt{n} - 1}} + \sum_{n = 10}^{\infty} \frac{1}{n^{\sqrt{n} - 1}} \\ \sum_{n = 10}^{\infty} \frac{1}{n^{\sqrt{n} - 1}} \leq \sum_{n = 10}^{\infty} \frac{1}{n^{2}} which converges ⟹ \sum_{n = 10}^{\infty} \frac{1}{n^{\sqrt{n} - 1}} also converges \\ ⟹ \sum_{n = 1}^{\infty} \frac{(- 1)^{n} n}{n^{\sqrt{n}}} converges absolutely \end{matrix}

3b

\begin{matrix} \sum_{n = 1}^{\infty} \frac{\cos (\frac{π n}{2})}{19 n + 7 \sqrt{n}} \\ Solution: \\ \cos (\frac{π n}{2}) = 0, - 1, 0, 1, 0, - 1, 0, 1, \dots \\ ⟹ S_{N} = 0, - 1, 0, 0, 0, - 1, 0, 0, 0, \dots \\ ⟹ | S_{N} | \leq 1 \\ ⟹ \sum_{n = 1}^{\infty} \cos (\frac{π n}{2}) is bounded \\ \frac{1}{19 n + 7 \sqrt{n}} is monotonically decreasing towards 0 \\ ⟹ By Dirichlet’s test: \sum_{n = 1}^{\infty} \frac{\cos (\frac{π n}{2})}{19 n + 7 \sqrt{n}} converges at least conditionally \\ \sum_{n = 1}^{\infty} | \frac{\cos (\frac{π n}{2})}{19 n + 7 \sqrt{n}} | = \sum_{n = 1}^{\infty} \frac{| \cos (\frac{π n}{2}) |}{19 n + 7 \sqrt{n}} \\ Let a_{n} = | \cos (\frac{π n}{2}) | \\ a_{2 n} = | \cos (π n) | = 1 \\ a_{2 n - 1} = | \cos (π n - \frac{π}{2}) | = 0 \\ 0 + 1 + 0 + 1 + \dots > \frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \dots ⟹ \sum a_{n} > \sum \frac{1}{3} \\ ⟹ \sum_{n = 1}^{\infty} \frac{| \cos (\frac{π n}{2}) |}{19 n + 7 \sqrt{n}} > \sum_{n = 1}^{\infty} \frac{1}{57 n + 21 \sqrt{n}} \underset{n > \sqrt{n}}{\underset{⏟}{>}} \sum_{n = 1}^{\infty} \frac{1}{78 n} which diverges \\ ⟹ \sum_{n = 1}^{\infty} \frac{| \cos (\frac{π n}{2}) |}{19 n + 7 \sqrt{n}} also diverges \\ ⟹ \sum_{n = 1}^{\infty} \frac{\cos (\frac{π n}{2})}{19 n + 7 \sqrt{n}} converges conditionally \end{matrix}

4a

\begin{matrix} Let f be a function \\ Let f be differentiable on R \\ Let {a_{n}}, {b_{n}} \subseteq R : a_{n} - b_{n} \to 0 \\ Prove or disprove: f (a_{n}) - f (b_{n}) \to 0 \\ Disproof: \\ Let f (x) = x^{2} \\ Let a_{n} = n + \frac{1}{n} \\ Let b_{n} = n \\ a_{n} - b_{n} = \frac{1}{n} \to 0 \\ f (a_{n}) - f (b_{n}) = {(n + \frac{1}{n})}^{2} - n^{2} = 2 + \frac{1}{n} \to 2 \\ ⟹ f (a_{n}) - f (b_{n}) ↛ 0 \end{matrix}

4b

\begin{matrix} Let f be a function \\ Let f be continuous on R \\ Let f has no local extremums \\ Prove or disprove: f is monotonic \\ Proof: \\ Let f be non-monotonic \\ ⟹ \exists a < b : f (a) > f (b), \exists c < d : f (c) < f (d) \\ Let a, b, c, d \in [α, β] \\ f is continuous ⟹ \exists x_{m}, x_{M} \in [α, β] : \forall x \in [α, β] : f (x_{m}) \leq f (x) \leq f (x_{M}) \\ f has no local extremums ⟹ {α, β} = {x_{m}, x_{M}} \\ Case 1. Let α = x_{m}, β = x_{M} \\ f (α) \leq f (b) < f (a) \leq f (β) ⟹ f (α) < f (β) \\ b \in [a, β], f (a) > f (b) ⟹ a is not a minimum of [a, β] \\ f (β) \geq f (a) > f (b) ⟹ β is not a minimum of [a, β] \\ ⟹ \exists x_{m} \in (a, β) : \forall x \in [a, β] : f (x) \geq f (x_{m}) \\ ⟹ x_{m} is a local minimum of f - Contradiction! \\ Case 2. Let β = x_{m}, α = x_{M} \\ f (β) \leq f (c) < f (d) \leq f (α) ⟹ f (β) < f (α) \\ c \in [α, d], f (c) < f (d) ⟹ d is not a minimum of [α, d] \\ f (α) \geq f (d) > f (c) ⟹ α is not a minimum of [α, d] \\ ⟹ \exists x_{m} \in (α, d) : \forall x \in [α, d] : f (x) \geq f (x_{m}) \\ ⟹ x_{m} is a local minimum of f - Contradiction! \\ ⟹ f is monotonic \\ Alternative proof: \\ Let f be non-monotonic \\ ⟹ \exists a < b : f (a) < f (b), \exists c < d : f (c) > f (d) \\ Let b \leq c (meaning that the non-monotonic part is a "hill") \\ Let f (b) \geq f (c) \\ ⟹ f (a) < f (b) > f (d) \\ a < b < c < d ⟹ b \in [a, d] \\ f is continuous ⟹ \exists x_{M} \in [a, d] : \forall x \in [a, d] : f (x) \leq f (x_{M}) \\ f (a) < f (b) ⟹ a \neq x_{M} \\ f (d) < f (b) ⟹ d \neq x_{M} \\ ⟹ x_{M} \in (a, d) ⟹ x_{M} is a local maximum of f - Contradiction! ⟹ f (b) \leq f (c) \\ ⟹ f (a) < f (c) > f (d) \\ a < b \leq c < d ⟹ c \in [a, d] \\ f is continuous ⟹ \exists x_{M} \in [a, d] : \forall x \in [a, d] : f (x) \leq f (x_{M}) \\ f (a) < f (c) ⟹ a \neq x_{M} \\ f (d) < f (c) ⟹ d \neq x_{M} \\ ⟹ x_{M} \in (a, d) ⟹ x_{M} is a local maximum of f - Contradiction! \\ Proof for non-monotonic part being a "pit" is similar \\ ⟹ f is monotonic \end{matrix}

5

\begin{matrix} Find for all a \in R : lim_{n \to \infty} (\sin (\sqrt{x - a}) - \sin \sqrt{x}) \\ Solution: \\ a = 0 ⟹ lim_{n \to \infty} (\sin (\sqrt{x}) - \sin (\sqrt{x})) = 0 \\ Let f (x) = \sin x \\ f is continuous and differentiable on R \\ By the Lagrange theorem: \exists c \in (\sqrt{x - a}, \sqrt{x}) : \frac{\sin (\sqrt{x}) - \sin (\sqrt{x - a})}{\sqrt{x} - \sqrt{x - a}} = f^{'} (c) = \cos (c) \\ ⟹ - 1 \leq \frac{\sin (\sqrt{x}) - \sin (\sqrt{x - a})}{\sqrt{x} - \sqrt{x - a}} \leq 1 \\ ⟹ \sqrt{x - a} - \sqrt{x} \leq \sin (\sqrt{x}) - \sin (\sqrt{x - a}) \leq \sqrt{x} - \sqrt{x - a} \\ lim_{x \to \infty} \sqrt{x} - \sqrt{x - a} = lim_{x \to \infty} \frac{a}{\underset{\to \infty}{\underset{⏟}{\sqrt{x}}} + \underset{\to \infty}{\underset{⏟}{\sqrt{x - a}}}} = 0 \\ ⟹ lim_{x \to \infty} \sqrt{x - a} - \sqrt{x} = - 0 = 0 \\ ⟹ - lim_{x \to \infty} \sin (\sqrt{x}) - \sin (\sqrt{x - a}) = lim_{n \to \infty} (\sin (\sqrt{x - a}) - \sin \sqrt{x}) = 0 \end{matrix}