Infi-1 6

1a

\begin{matrix} a_{n} = \frac{1}{n} + \frac{1}{n + 1} + \dots + \frac{1}{3 n} \\ Prove: lim_{n \to \infty} a_{n} = L \geq \frac{2}{3} \\ Proof: \\ a_{n + 1} - a_{n} = \frac{1}{3 n + 3} + \frac{1}{3 n + 2} + \frac{1}{3 n + 1} - \frac{1}{n} < \frac{3}{3 n} - \frac{1}{n} = 0 \\ ⟹ a_{n} - monotonically decreasing \\ \frac{1}{3 n} \leq \frac{1}{3 n - 1} \leq \dots \leq \frac{1}{n + 1} \leq \frac{1}{n} \\ ⟹ \frac{2 n + 1}{3 n} \leq \underset{\geq \frac{1}{3 n}}{\underset{⏟}{\frac{1}{n}}} + \underset{\geq \frac{1}{3 n}}{\underset{⏟}{\frac{1}{n + 1}}} + \dots + \frac{1}{3 n} \\ \frac{2}{3} = \frac{2 n}{3 n} \leq \frac{2 n + 1}{3 n} \leq a_{n} \\ ⟹ a_{n} is lower-bounded by \frac{2}{3} ⟹ lim_{n \to \infty} a_{n} = L \geq \frac{2}{3} \end{matrix}

1b

\begin{matrix} Explain in short why a_{n} from 1a does not converge to 0 as per the limit arithmetics \\ Explanation: \\ As n grows towards infinity, number of terms in the sum is also growing towards infinity, \\ even faster than n does itself, and thus the limit arithmetics do not apply in this case \end{matrix}

1c

\begin{matrix} a_{n} \to L \\ 1. Prove or disprove: \forall n : a_{n} > x ⟹ lim_{n \to \infty} a_{n} > x \\ 2. Prove or disprove: \forall n : a_{n} > x ⟹ lim_{n \to \infty} a_{n} \geq x \\ Disproof for 1. \\ Let a_{n} = \frac{1}{n} \\ \forall n : a_{n} > 0 \\ lim_{n \to \infty} a_{n} = 0 ≯ 0 \\ ⟹ \forall n : a_{n} > x ⟹ ⧸ lim_{n \to \infty} a_{n} > x \\ Proof for 2. \\ Let lim_{n \to \infty} a_{n} = L < x \\ ⟹ x - L > 0 \\ a_{n} > x ⟹ a_{n} - x > 0 \\ Let ε = \frac{x - L}{2} \\ \exists N : \forall n > N : | a_{n} - L | < ε \\ | a_{n} - L | = | (a_{n} - x) + (x - L) | \leq | a_{n} - x | + | x - L | < ε \\ ⟹ (a_{n} - x) + (x - L) < \frac{x - L}{2} \\ ⟹ \underset{> 0}{\underset{⏟}{(a_{n} - x)}} + \underset{> 0}{\underset{⏟}{\frac{x - L}{2}}} < 0 - Contradiction! \\ ⟹ lim_{n \to \infty} a_{n} \geq x ⟹ \forall n : a_{n} > x ⟹ lim_{n \to \infty} a_{n} \geq x \end{matrix}

2

\begin{matrix} a_{n} : \sqrt{2}, \sqrt{2 + \sqrt{2}}, \sqrt{2 + \sqrt{2 + \sqrt{2}}}, \dots \end{matrix}

2a

\begin{matrix} Write a_{n} in the form or recursive formula: \\ a_{n} = {\begin{cases} \sqrt{2} & n = 1 \\ \sqrt{2 + a_{n - 1}} & otherwise \end{cases} \end{matrix}

2b

\begin{matrix} Prove a_{n} is upper-bounded and monotonically non-decreasing, find lim_{n \to \infty} a_{n} \\ Proof: \\ Base case. a_{1} = \sqrt{2} \\ 0 < \sqrt{2} < 2 ⟹ 0 < a_{1} < 2 \\ Induction step. Let 0 < a_{n} < 2 \\ a_{n + 1} = \sqrt{2 + a_{n}} < \sqrt{2 + 2} = \sqrt{4} = 2 \\ ⟹ a_{n + 1} < 2 \\ \sqrt{2 + a_{n}} > \sqrt{2} > 0 ⟹ a_{n + 1} > 0 \\ ⟹ \forall n \in N : 0 < a_{n} < 2 & (1) \\ a_{n + 1} \geq a_{n} ⟺ \sqrt{2 + a_{n}} \geq a_{n} ⟺ 2 + a_{n} \geq a_{n}^{2} \\ ⟺ a_{n}^{2} - a_{n} - 2 < 0 ⟺ - 1 < a_{n} < 2 \\ \forall n \in N : 0 < a_{n} < 2 ⟹ - 1 < a_{n} < 2 ⟹ a_{n + 1} \geq a_{n} \\ ⟹ a_{n} is monotonically non-decreasing & (2) \\ (1) \land (2) ⟹ \exists lim_{n \to \infty} a_{n} = L \in R \\ Let lim_{n \to \infty} a_{n} = L \\ L = lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} \sqrt{2 + a_{n}} = \sqrt{2 + L} \\ L = \sqrt{2 + L} ⟹ L^{2} = 2 + L ⟹ L^{2} - L - 2 = 0 \\ ⟹ (L - 2) (L + 1) = 0 ⟹ [\begin{matrix} L = 2 \\ L = - 1 \end{matrix} \\ \forall n \in N : a_{n} > 0 ⟹ lim_{n \to \infty} a_{n} \geq 0 ⟹ L \geq 0 ⟹ L = 2 \\ ⟹ lim_{n \to \infty} a_{n} = 2 \end{matrix}

3

\begin{matrix} a_{n} = {\begin{cases} 1 & n = 1 \\ \sqrt{3 a_{n - 1}} & otherwise \end{cases} \\ Prove a_{n} converges and find lim_{n \to \infty} a_{n} \\ Proof: \\ Base case. a_{1} = 1 \\ 1 \leq 1 < 3 ⟹ 1 \leq a_{1} < 3 \\ Induction step. Let 1 \leq a_{n} < 3 \\ a_{n + 1} = \sqrt{3 a_{n}} < \sqrt{3 \cdot 3} = 3 ⟹ a_{n + 1} < 3 \\ \sqrt{3 a_{n}} \geq \sqrt{3} \geq 1 ⟹ a_{n + 1} \geq 1 \\ ⟹ \forall n \in N : 1 \leq a_{n} < 3 & (1) \\ a_{n + 1} > a_{n} ⟺ \sqrt{3 a_{n}} > a_{n} ⟺ 3 a_{n} > a_{n}^{2} \\ ⟺ a_{n} (a_{n} - 3) < 0 ⟺ 0 < a_{n} < 3 \\ \forall n > 1 \in N : 1 \leq a_{n} < 3 ⟹ 0 < a_{n} < 3 ⟹ a_{n + 1} > a_{n} \\ ⟹ a_{n} is monotonically increasing & (2) \\ (1) \land (2) ⟹ \exists lim_{n \to \infty} a_{n} = L \in R \\ Let lim_{n \to \infty} a_{n} = L \\ L = lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} \sqrt{3 a_{n}} = \sqrt{3 L} \\ L = \sqrt{3 L} ⟺ L^{2} = 3 L ⟺ [\begin{matrix} L = 0 \\ L = 3 \end{matrix} \\ \forall n \in N : a_{n} \geq 1 ⟹ a_{n} > \frac{1}{2} ⟹ lim_{n \to \infty} a_{n} \geq \frac{1}{2} ⟹ lim_{n \to \infty} \neq 0 \\ ⟹ lim_{n \to \infty} a_{n} = 3 \end{matrix}

4

\begin{matrix} a_{1} > 0 \\ a_{n + 1} = a_{n} \cdot e^{a_{n}} \\ Find lim_{n \to \infty} a_{n} \\ Solution: \\ Base case. a_{1} > 0 \\ Induction step. Let a_{n} > 0 \\ a_{n + 1} = \underset{> 0}{\underset{⏟}{a_{n}}} \cdot \underset{> 0}{\underset{⏟}{e^{a_{n}}}} > 0 ⟹ a_{n + 1} > 0 \\ ⟹ \forall n \in N : a_{n} > 0 \\ a_{n + 1} > a_{n} ⟺ a_{n} \cdot e^{a_{n}} > a_{n} ⟺ e^{a_{n}} > 1 ⟺ a_{n} > 0 \\ \forall n \in N : a_{n} > 0 ⟹ a_{n + 1} > a_{n} \\ ⟹ a_{n} is monotonically increasing \\ Let lim_{n \to \infty} a_{n} = L \in R \\ L = lim_{n \to \infty} a_{n + 1} = a_{n} \cdot e^{a_{n}} = L \cdot e^{L} \\ L = L \cdot e^{L} ⟺ L (e^{L} - 1) = 0 ⟺ [\begin{matrix} L = 0 \\ e^{L} = 1 \end{matrix} ⟺ L = 0 \\ a_{n} is monotonocally increasing ⟹ \exists lim_{n \to \infty} a_{n} \\ L \geq a_{1} > 0 ⟹ L \notin R ⟹ lim_{n \to \infty} a_{n} = \infty \end{matrix}

5a

\begin{matrix} Let a_{n} be a monotonically non-decreasing sequence \\ Let C \in R \\ Prove: a_{n} + C is monotonically non-decreasing \\ Proof: \\ Let b_{n} = a_{n} + C \\ \forall n : a_{n + 1} \geq a_{n} ⟹ \forall n : a_{n + 1} + C \geq a_{n} + C ⟹ b_{n + 1} \geq b_{n} \\ ⟹ b_{n} is monotonically non-decreasing ⟹ a_{n} + C is monotonically non-decreasing \end{matrix}

5b

\begin{matrix} Let a_{n} be a monotonically non-decreasing sequence \\ Let C \in R \\ Prove: C > 0 ⟹ a_{n} \cdot C is monotonically non-decreasing \\ Prove: C < 0 ⟹ a_{n} \cdot C is monotonically non-increasing \\ Proof: \\ Let b_{n} = a_{n} \cdot C, C > 0 \\ \forall n : a_{n + 1} \geq a_{n} \underset{C > 0}{⟹} a_{n + 1} \cdot C \geq a_{n} \cdot C ⟹ b_{n + 1} \geq b_{n} \\ ⟹ b_{n} is monotonically non-decreasing \\ ⟹ C > 0 ⟹ a_{n} \cdot C is monotonically non-decreasing \\ Let c_{n} = a_{n} \cdot C, C < 0 \\ \forall n : a_{n + 1} \geq a_{n} \underset{C < 0}{⟹} a_{n + 1} \cdot C \leq a_{n} \cdot C ⟹ b_{n + 1} \leq b_{n} \\ ⟹ c_{n} is monotonically non-increasing \\ ⟹ C < 0 ⟹ a_{n} \cdot C is monotonically non-increasing \end{matrix}

5c

\begin{matrix} Let a_{n} be a monotonically non-decreasing sequence \\ a_{n} > 0 \\ Prove or disprove: \frac{1}{a_{n}} is monotonically non-increasing \\ Proof: \\ Let b_{n} = \frac{1}{a_{n}} \\ \forall n : a_{n + 1} \geq a_{n} \underset{a_{n} > 0}{⟹} \forall n : \frac{1}{a_{n + 1}} \leq \frac{1}{a_{n}} ⟹ b_{n + 1} \leq b_{n} \\ ⟹ b_{n} is monotonically non-increasing ⟹ \frac{1}{a_{n}} is monotonically non-increasing \end{matrix}

5d

\begin{matrix} Let a_{n} be a monotonically non-decreasing sequence \\ a_{n} \neq 0 \\ Prove or disprove: \frac{1}{a_{n}} is monotonically non-increasing \\ Disproof: \\ Let a_{1} = - 1, a_{n} = n \\ \frac{1}{a_{n}} : - 1, \frac{1}{2}, \frac{1}{3}, \dots \\ \frac{1}{a_{1}} < \frac{1}{a_{2}} > \frac{1}{a_{3}} ⟹ \frac{1}{a_{n}} is not monotonically non-increasing \end{matrix}

5e

\begin{matrix} Let a_{n} be a monotonically non-decreasing sequence \\ a_{n} \neq 0 \\ b_{n} = \frac{1}{a_{n}} \\ Prove or disprove: \exists N : \forall n > N : b_{n + 1} \leq b_{n} \\ Proof: \\ Let \exists N : \forall n > N : a_{n} < 0 \\ a_{n + 1} \geq a_{n} \underset{a_{n} < 0}{⟹} \frac{1}{a_{n + 1}} \leq \frac{1}{a_{n}} ⟹ \exists N : \forall n > N : b_{n + 1} \leq b_{n} \\ Let ∄ N : \forall n > N : a_{n} < 0 \\ ⟹ \forall n : a_{n} > 0 \\ a_{n + 1} \geq a_{n} ⟹ \frac{1}{a_{n + 1}} \leq \frac{1}{a_{n}} ⟹ b_{n + 1} \leq b_{n} ⟹ b_{n} is monotonically non-increasing \\ ⟹ \exists N = 1 : \forall n > N : b_{n + 1} \leq b_{n} \end{matrix}

5f

\begin{matrix} Let a_{n}, b_{n} be a monotonically non-decreasing sequences \\ Prove or disprove: a_{n} + b_{n} is monotonically non-decreasing \\ Proof: \\ \forall n : a_{n + 1} \geq a_{n}, b_{n + 1} \geq b_{n} ⟹ a_{n + 1} + b_{n + 1} \geq a_{n} + b_{n} \\ ⟹ a_{n} + b_{n} is monotonically non-decreasing \end{matrix}

5g

\begin{matrix} Let a_{n}, b_{n} be a monotonically non-decreasing sequences \\ a_{n} > 0, b_{n} > 0 \\ Prove or disprove: a_{n} b_{n} is monotonically non-decreasing \\ Proof: \\ \forall n : a_{n + 1} \geq a_{n}, b_{n + 1} \geq b_{n} \\ a_{n} > 0, b_{n} > 0 ⟹ a_{n} b_{n} > 0 \\ a_{n + 1} b_{n + 1} \underset{b_{n + 1} = b_{n} + x, x \geq 0}{\geq} a_{n + 1} b_{n} \underset{a_{n + 1} = a_{n} + y, y \geq 0}{\geq} a_{n} b_{n} \\ ⟹ a_{n} b_{n} is monotonically non-decreasing \end{matrix}

5h

\begin{matrix} Let a_{n}, b_{n} be monotonic \\ a_{n} > 0, b_{n} > 0 \\ Prove or disprove: a_{n} + b_{n} is monotonic \\ Disproof: \\ a_{n} = n \\ b_{n} : 3, 1, 1, \dots \\ c_{n} = a_{n} + b_{n} = 4, 3, 4, 5, \dots \\ c_{1} > c_{2} < c_{3} \\ ⟹ a_{n} + b_{n} is not monotonic \end{matrix}

6a

\begin{matrix} a_{n} = (- 1)^{n + 1} \cdot \frac{n + 12}{n} \\ Find lim^{―} a_{n}, lim_{―} a_{n} \\ Solution: \\ a_{2 n} = (- 1) \cdot \frac{n + 6}{n} \to - 1 \\ a_{2 n - 1} = 1 \cdot \frac{2 n + 11}{2 n - 1} \to 1 \\ All elements of a_{n} are either in a_{2 n} or in a_{2 n - 1} \\ ⟹ {- 1, 1} is a set of partial limits of a_{n} \\ ⟹ lim^{―} a_{n} = 1, lim_{―} a_{n} = - 1 \end{matrix}

6b

\begin{matrix} b_{n} = \cos (\frac{n π}{3}) \\ Find lim^{―} b_{n}, lim_{―} b_{n} \\ Solution: \\ Exist many partial limits of b_{n} \\ But they are all between (- 1) and 1 \\ m a x (b_{n}) = 1, b_{6 n} = \cos (2 n π) = 1 \to 1 ⟹ lim^{―} b_{n} = 1 \\ m i n (b_{n}) = - 1, b_{9 n} = \cos (3 n π) = - 1 \to - 1 ⟹ lim_{―} b_{n} = - 1 \end{matrix}

6c

\begin{matrix} c_{n} = (- 2)^{n} - 4 \\ Find lim^{―} a_{n}, lim_{―} a_{n} \\ Solution: \\ c_{2 n} = 2^{2 n} - 4 \to \infty \\ c_{2 n - 1} = - 2^{2 n - 1} - 4 \to - \infty \\ All elements of c_{n} are either in c_{2 n} or in c_{2 n - 1} \\ ⟹ {- \infty, \infty} is a set of partial limits of c_{n} \\ ⟹ lim^{―} a_{n} = \infty, lim_{―} a_{n} = - \infty \end{matrix}