Infi-1 3

1a

\begin{matrix} a_{n} = \frac{1}{n + 27} \\ Prove: lim_{n \to \infty} a_{n} = 0 \\ Proof: \\ | \frac{1}{n + 27} - 0 | = \frac{1}{n + 27} \underset{n > N}{<} \frac{1}{N + 27} \\ Let N = \frac{1}{ε}, ε > 0 \\ | \frac{1}{n + 27} - 0 | \underset{n > N}{<} \frac{1}{\frac{1}{ε} + 27} < \frac{1}{\frac{1}{ε}} = ε \\ ⟹ \forall ε > 0 : \exists N : \forall n > N, n \in N : | \frac{1}{n + 27} - 0 | < ε \\ ⟺ lim_{n \to \infty} a_{n} = 0 \end{matrix}

1b

\begin{matrix} a_{n} = \frac{n + 3}{n + 32} \\ Prove: lim_{n \to \infty} a_{n} = 1 \\ Proof: \\ | \frac{n + 3}{n + 32} - 1 | = | \frac{n + 3 - (n + 32)}{n + 32} | = \frac{29}{n + 32} \underset{n > N}{<} \frac{29}{N + 32} \\ Let N = \frac{29}{ε}, ε > 0 \\ | \frac{n + 3}{n + 32} - 1 | \underset{n > N}{<} \frac{29}{\frac{29}{ε} + 32} < \frac{29}{\frac{29}{ε}} = ε \\ ⟹ \forall ε > 0 : \exists N : \forall n > N, n \in N : | \frac{n + 3}{n + 32} - 1 | < ε \\ ⟺ lim_{n \to \infty} a_{n} = 1 \end{matrix}

1c

\begin{matrix} a_{n} = \frac{4 n^{2} - 25}{n^{2} - 16} \\ Prove: lim_{n \to \infty} a_{n} = 4 \\ Proof: \\ | \frac{4 n^{2} - 25}{n^{2} - 16} - 4 | = | \frac{4 n^{2} - 25 - 4 (n^{2} - 16)}{n^{2} - 16} | = \frac{39}{n^{2} - 16} \underset{n > N}{<} \frac{39}{n^{2} - 16} \\ Let N = \sqrt{\frac{39}{ε} + 16} \\ | \frac{4 n^{2} - 25}{n^{2} - 16} - 4 | \underset{n > N}{<} \frac{39}{{\sqrt{\frac{39}{ε} + 16}}^{2} - 16} = \frac{39}{\frac{39}{ε} + 16 - 16} = \frac{39}{\frac{39}{ε}} = ε \\ ⟹ \forall ε > 0 : \exists N : \forall n > N, n \in N : | \frac{4 n^{2} - 25}{n^{2} - 16} - 4 | < ε \\ ⟺ lim_{n \to \infty} a_{n} = 4 \end{matrix}

2a

\begin{matrix} a_{n} = \frac{(- 1)^{n}}{n^{8}} \\ Find lim_{n \to \infty} a_{n} \\ Solution: \\ Let’s assume lim_{n \to \infty} a_{n} = 0 \\ | \frac{(- 1)^{n}}{n^{8}} - 0 | = \frac{1}{n^{8}} \underset{n > N}{<} \frac{1}{N^{8}} \\ Let N = \sqrt[8]{\frac{1}{ε}} \\ | \frac{(- 1)^{n}}{n^{8}} - 0 | < \frac{1}{{\sqrt[8]{\frac{1}{ε}}}^{8}} = \frac{1}{\frac{1}{ε}} = ε \\ ⟹ \forall ε > 0 : \exists N : \forall n > N, n \in N : | \frac{(- 1)^{n}}{n^{8}} - 0 | < ε \\ ⟺ lim_{n \to \infty} a_{n} = 0 \end{matrix}

2b

\begin{matrix} a_{n} = \frac{1 + (- 1)^{n}}{2^{n} + 24} \\ Find lim_{n \to \infty} a_{n} \\ Solution: \\ Let’s assume lim_{n \to \infty} a_{n} = 0 \\ | \frac{1 + (- 1)^{n}}{2^{n} + 24} - 0 | = \frac{1 + (- 1)^{n}}{2^{n} + 24} \leq \frac{2}{2^{n} + 24} \underset{n > N}{<} \frac{2}{2^{N} + 24} \\ Let N = 1 + \log_{2} (\frac{1}{ε}), ε > 0 \\ | \frac{1 + (- 1)^{n}}{2^{n} + 24} - 0 | \underset{n > N}{<} \frac{2}{2^{1 + \log_{2} (1 / ε)} + 24} = \frac{2}{2 \frac{1}{ε} + 24} < \frac{2}{2 \frac{1}{ε}} = ε \\ ⟹ \forall ε > 0 : \exists N : \forall n > N, n \in N : | \frac{1 + (- 1)^{n}}{2^{n} + 24} - 0 | < ε \\ ⟺ lim_{n \to \infty} a_{n} = 0 \end{matrix}

2c

\begin{matrix} a_{n} = \frac{n^{3} - n^{2} + \sqrt{n}}{n^{3} + n^{2} - \sqrt{n}} \\ Find lim_{n \to \infty} a_{n} \\ Solution: \\ Let’s assume lim_{n \to \infty} a_{n} = 1 \\ | \frac{n^{3} - n^{2} + \sqrt{n}}{n^{3} + n^{2} - \sqrt{n}} - 1 | = | \frac{n^{3} - n^{2} + \sqrt{n} - (n^{3} + n^{2} - \sqrt{n})}{n^{3} + n^{2} - \sqrt{n}} | = | \frac{2 (\sqrt{n} - n^{2})}{n^{3} + n^{2} - \sqrt{n}} | = \\ \underset{n^{2} > \sqrt{n}}{=} \frac{2 (n^{2} - \sqrt{n})}{n^{3} + n^{2} - \sqrt{n}} < \frac{2 n^{2}}{n^{3} + n^{2} - \sqrt{n}} \underset{n^{2} - \sqrt{n} > 0}{<} \frac{2 n^{2}}{n^{3}} = \frac{2}{n} \underset{n > N}{<} \frac{2}{N} \\ Let N = \frac{2}{ε}, ε > 0 \\ | \frac{n^{3} - n^{2} + \sqrt{n}}{n^{3} + n^{2} - \sqrt{n}} - 1 | < \frac{2}{\frac{2}{ε}} = ε \\ ⟹ \forall ε > 0 : \exists N : \forall n > N, n \in N : | \frac{n^{3} - n^{2} + \sqrt{n}}{n^{3} + n^{2} - \sqrt{n}} - 1 | < ε \\ ⟺ lim_{n \to \infty} a_{n} = 1 \end{matrix}

3

\begin{matrix} Prove: lim_{n \to \infty} \frac{\sqrt{n} + n}{2 n - 7 \sqrt{n}} \neq \frac{1}{7} \\ Proof: \\ | \frac{\sqrt{n} + n}{2 n - 7 \sqrt{n}} - \frac{1}{7} | = | \frac{\overset{7 \sqrt{n} + 7 n - 2 n + 7 \sqrt{n} = 5 n + 14 \sqrt{n}}{\overset{⏞}{7 (\sqrt{n} + n) - (2 n - 7 \sqrt{n})}}}{7 (2 n - 7 \sqrt{n})} | = \frac{5 n + 14 \sqrt{n}}{14 n - 49 \sqrt{n}} > \\ > \frac{5 n}{14 n - 49 \sqrt{n}} > \frac{5 n}{14 n} = \frac{5}{14} \\ ⟹ \exists ε > 0 : \forall n \in N : | \frac{\sqrt{n} + n}{2 n - 7 \sqrt{n}} - \frac{1}{7} | \geq ε \\ ⟺ lim_{n \to \infty} \frac{\sqrt{n} + n}{2 n - 7 \sqrt{n}} \neq \frac{1}{7} \end{matrix}

4a

\begin{matrix} Show that \\ \forall a, b, c \in R : (| a - b | < 7) \land (| b - c | < 48) ⟹ | a - c | < 55 \\ Solution: \\ | a - c | = | a - b + b - c | \leq \underset{< 7}{\underset{⏟}{| a - b |}} + | b - c | < 7 + \underset{< 48}{\underset{⏟}{| b - c |}} < 7 + 48 = 55 \\ | a - c | < 55 \end{matrix}

4b

\begin{matrix} Show that \forall a, b, c, ϵ, ζ \in R, ϵ > 0, ζ > 0 : \\ (| a - b | < ϵ) \land (| b - c | < ζ) ⟹ | a - c | < ϵ + ζ \\ Solution: \\ | a - c | = | a - b + b - c | \leq \underset{< ϵ}{\underset{⏟}{| a - b |}} + | b - c | < ϵ + \underset{< ζ}{\underset{⏟}{| b - c |}} < ϵ + ζ \\ | a - c | < ϵ + ζ \end{matrix}

Given: lim_{n \to \infty} a_{n} = L ⟺ \forall ε > 0 \exists N \in N : \forall n \in N, n > N : | a_{n} - L | < ε

5a

\begin{matrix} Show that definition (1) \\ lim_{n \to \infty} a_{n} = L_{1} ⟺ \forall ε > 0 \exists N \in N : \forall n \in N, n > N : | a_{n} - L_{1} | < ε \\ and definition (2) \\ lim_{n \to \infty} a_{n} = L_{2} ⟺ \forall ε > 0 \exists N \in N : \forall n \in N, n \geq N : | a_{n} - L_{2} | < ε \\ are equivalent and L_{1} = L_{2} \\ Solution: \\ Let the definition (1) be true \\ Let lim_{n \to \infty} a_{n} = L_{1} \\ Then \forall ε > 0 \exists N_{1} \in N : \forall n \in N, n > N_{1} : | a_{n} - L_{1} | < ε \\ Let N_{2} = N_{1} + 1 \\ n > N_{1} ⟺ n > N_{2} - 1 \\ {\begin{matrix} n = N_{2} : n = N_{2} > N_{2} - 1 \\ n > N_{2} : n > N_{2} > N_{2} - 1 \end{matrix} ⟹ n > N_{1} ⟺ n \geq N_{2} \\ (\forall ε > 0 \exists N_{1} \in N : \forall n \in N, n > N_{1} : | a_{n} - L_{1} | < ε) \\ ⟹ (\forall ε > 0 \exists N_{2} \in N : \forall n \in N, n \geq N_{2} : | a_{n} - L_{1} | < ε) \\ ⟹ (1) ⟹ (2) [1] \\ Now let the definition (2) be true \\ Let lim_{n \to \infty} a_{n} = L_{2} \\ Then \forall ε > 0 \exists N_{1} \in N : \forall n \in N, n \geq N_{1} : | a_{n} - L_{2} | < ε \\ Let N_{2} = N_{1} - 1 \\ n \geq N_{1} ⟺ n \geq N_{2} + 1 > N_{2} ⟹ n \geq N_{1} ⟺ n > N_{2} \\ (\forall ε > 0 \exists N_{1} \in N : \forall n \in N, n \geq N_{1} : | a_{n} - L_{2} | < ε) \\ ⟹ (\forall ε > 0 \exists N_{2} \in N : \forall n \in N, n > N_{2} : | a_{n} - L_{1} | < ε) \\ ⟹ (2) ⟹ (1) [2] \\ [1] and [2] ⟹ (1) \equiv (2) \\ Let L_{1} \neq L_{2} \\ Then ζ = | L_{1} - L_{2} | > 0, Let ε = \frac{ζ}{4} \\ | L_{1} - L_{2} | \leq | L_{1} - a_{n} | + | a_{n} - L_{2} | < 2 ε = \frac{ζ}{2} \\ ⟹ ζ < \frac{ζ}{2} - Contradiction! ⟹ L_{1} = L_{2} \end{matrix}

5b

\begin{matrix} Show that definition (1) \\ lim_{n \to \infty} a_{n} = L ⟺ \forall ε > 0 \exists N \in N : \forall n \in N, n > N : | a_{n} - L | < ε \\ and definition (2) \\ lim_{n \to \infty} a_{n} = L ⟺ \forall ε > 0 \exists N \in N : \forall n \in N, n > N : | a_{n} - L | \leq ε \\ are equivalent \\ Solution: \\ One direction is obvious, as \forall r \in R : r < ε ⟹ r \leq ε \\ ⟹ (1) ⟹ (2) [1] \\ Now let the definition (2) be true \\ Let lim_{n \to \infty} a_{n} = L \\ Then \forall ε > 0 \exists N \in N : \forall n \in N, n \geq N : | a_{n} - L | \leq ε \\ Let ε_{1} = \frac{ε}{2}, ε > 0 \\ \forall r \in R : r \leq ε_{1} ⟹ r \leq \frac{ε}{2} < ε \\ (\forall ε_{1} > 0 \exists N_{1} \in N : \forall n \in N, n > N : | a_{n} | \leq ε_{1}) \\ ⟹ (\forall ε > 0 \exists N_{2} \in N : \forall n \in N, n > N_{2} : | a_{n} - L_{1} | < ε) \\ ⟹ (2) ⟹ (1) [2] \\ [1] and [2] ⟹ (1) \equiv (2) \end{matrix}

5c

\begin{matrix} Show that definition (1) \\ lim_{n \to \infty} a_{n} = L ⟺ \forall ε > 0 \exists N \in N : \forall n \in N, n > N : | a_{n} - L | < ε \\ and definition (2) \\ lim_{n \to \infty} a_{n} = L ⟺ \forall ε \geq 0 \exists N \in N : \forall n \in N, n > N : | a_{n} - L | < ε \\ are not equivalent \\ Solution: \\ 1. \forall r \in R : | r | \geq 0 \\ ε = 0 ⟹ | a_{n} - L | ≮ ε = 0 \\ 2. Let us replace | a_{n} - L | < ε with | a_{n} - L | \leq ε as per 5b \\ This wouldn’t help either, for example: \\ (1) ⟹ lim_{n \to \infty} \frac{1}{n} = 0 \\ But \forall n \in N : | \frac{1}{n} | ≰ 0 ⟹ (2) ⟹ ⧸ lim_{n \to \infty} \frac{1}{n} = 0 \\ ⟹ (1) ≢ (2) \end{matrix}

6

\begin{matrix} Given: lim_{n \to \infty} a_{n} = π \\ b_{n} = {\begin{cases} 2 a_{n} & n = 314 \\ a_{n} & otherwise \end{cases} \\ Prove: lim_{n \to \infty} b_{n} = π \\ Proof: \\ lim_{n \to \infty} a_{n} = π ⟺ \forall ε > 0 : \exists N : \forall n \in N, n > N : | a_{n} - π | < ε \\ Let N_{1} = m a x (N, 314) \\ \forall n \in N, n > N_{1} : b_{n} = a_{n} \\ \forall ε > 0 : \exists N_{1} : \forall n \in N, n > N_{1} : | a_{n} - π | < ε \\ ⟹ \forall ε > 0 : \exists N_{1} : \forall n \in N, n > N_{1} : | b_{n} - π | < ε ⟺ lim_{n \to \infty} b_{n} = π \end{matrix}