Infi-1 4

1

\begin{matrix} Given: a_{n} \to L \in R, x \neq L \\ Prove: \exists N : \forall n > N : a_{n} \neq x \\ Proof: \\ lim_{n \to \infty} a_{n} = L ⟺ \forall ε > 0 : \exists N : \forall n > N : | a_{n} - L | < ε \\ x \neq L ⟹ | x - L | > 0 \\ Let ε = | x - L | \\ \exists N : \forall n > N : | a_{n} - L | < ε \\ Let x > L : | a_{n} - L | < x - L ⟺ 2 L - x < a_{n} < x ⟹ a_{n} < x ⟹ a_{n} \neq x \\ Let x < L : | a_{n} - L | < L - x ⟺ x < a_{n} < 2 L - x ⟹ a_{n} > x ⟹ a_{n} \neq x \\ ⟹ \exists N : \forall n > N : a_{n} \neq x \end{matrix}

2a

\begin{matrix} Prove: lim_{n \to \infty} a_{n} = L > 0 ⟹ lim_{n \to \infty} \sqrt[n]{a_{n}} = 1 \\ Proof: \\ Let ε = \frac{L}{4} \\ | a_{n} - L | < \frac{L}{4} ⟺ \frac{3 L}{4} < a_{n} < \frac{5 L}{4} \\ ⟺ \underset{\to 1}{\underset{⏟}{\sqrt[n]{\frac{3 L}{4}}}} < \sqrt[n]{a_{n}} < \underset{\to 1}{\underset{⏟}{\sqrt[n]{\frac{5 L}{4}}}} \\ By the "Sandwich" theorem: lim_{n \to \infty} \sqrt[n]{a_{n}} = 1 \end{matrix}

2b

\begin{matrix} Prove or disprove: If a_{n} does not converge to L > 0, \sqrt[n]{a_{n}} still converges to 1 \\ Disproof: \\ a_{n} = (- 1)^{n} ⟹ ∄ lim_{n \to \infty} a_{n} \\ \sqrt[n]{a_{n}} = \sqrt[n]{(- 1)^{n}} = - 1 ⟹ lim_{n \to \infty} \sqrt[n]{a_{n}} = - 1 \neq 1 \end{matrix}

3a

\begin{matrix} a_{n} = \frac{2 n^{5} + 16 n^{2} + 4}{8 n^{5} - 24 n^{4} - 3} \\ lim_{n \to \infty} a_{n} = lim_{n \to \infty} n^{5} \frac{2 + \frac{16}{n^{3}} + \frac{4}{n^{5}}}{8 n^{5} - 24 n^{4} - 3} = lim_{n \to \infty} \frac{2 + \overset{\to 0}{\overset{⏞}{\frac{16}{n^{3}}}} + \overset{\to 0}{\overset{⏞}{\frac{4}{n^{5}}}}}{8 - \underset{\to 0}{\underset{⏟}{\frac{24}{n}}} - \underset{\to 0}{\underset{⏟}{\frac{3}{n^{5}}}}} = \frac{2}{8} = \frac{1}{4} \\ lim_{n \to \infty} a_{n} = \frac{1}{4} \end{matrix}

3b

\begin{matrix} b_{n} = \frac{n^{3} - n^{2} + n - \sqrt{n} + 1}{2 n^{3} + 12 \sqrt{n} - 5} \\ lim_{n \to \infty} b_{n} = lim_{n \to \infty} n^{3} \frac{1 - \frac{1}{n} + \frac{1}{n^{2}} - \sqrt{\frac{1}{n^{5}}} + \frac{1}{n^{3}}}{2 n^{3} + 12 \sqrt{n} - 5} = lim_{n \to \infty} \frac{1 - \overset{\to 0}{\overset{⏞}{\frac{1}{n}}} + \overset{\to 0}{\overset{⏞}{\frac{1}{n^{2}}}} - \overset{\to 0}{\overset{⏞}{\sqrt{\frac{1}{n^{5}}}}} + \overset{\to 0}{\overset{⏞}{\frac{1}{n^{3}}}}}{2 + \underset{\to 0}{\underset{⏟}{12 \sqrt{\frac{1}{n^{5}}}}} - \underset{\to 0}{\underset{⏟}{\frac{5}{n^{3}}}}} = \frac{1}{2} \\ lim_{n \to \infty} b_{n} = \frac{1}{2} \end{matrix}

3c

\begin{matrix} c_{n} = \frac{\sqrt{n + 2} - \sqrt{n + 1}}{\sqrt{n}} = \frac{(\sqrt{n + 2} - \sqrt{n + 1}) (\sqrt{n + 2} + \sqrt{(n + 1)})}{\sqrt{n} (\sqrt{n + 2} + \sqrt{n + 1})} = \\ = \frac{n + 2 - (n + 1)}{\sqrt{n} (\sqrt{n + 2} + \sqrt{n + 1})} = \frac{1}{\sqrt{n} (\sqrt{n + 2} + \sqrt{n + 1})} = \\ = \frac{1}{\sqrt{n^{2} + 2 n} + \sqrt{n^{2} + n}} = \frac{1}{n (\sqrt{1 + \frac{2}{n}} + \sqrt{1 + \frac{1}{n}})} \\ lim_{n \to \infty} c_{n} = lim_{n \to \infty} \frac{1}{n (\sqrt{1 + \frac{2}{n}} + \sqrt{1 + \frac{1}{n}})} = lim_{n \to \infty} \underset{\to 0}{\underset{⏟}{\frac{1}{n}}} \cdot \underset{\to \frac{1}{2}}{\underset{⏟}{\frac{1}{\underset{\to 1}{\underset{⏟}{\sqrt{1 + \frac{2}{n}}}} + \underset{\to 1}{\underset{⏟}{\sqrt{1 + \frac{1}{n}}}}}}} = 0 \\ lim_{n \to \infty} c_{n} = 0 \end{matrix}

3d

\begin{matrix} d_{n} = \frac{\sqrt[3]{n + 2} - \sqrt[3]{n + 1}}{\sqrt[3]{n}} = \\ = \frac{(\sqrt[3]{n + 2} - \sqrt[3]{n + 1}) (\sqrt[3]{(n + 2)^{2}} + \sqrt[3]{(n + 2) (n + 1)} + \sqrt[3]{(n + 1)^{2}})}{\sqrt[3]{n} (\sqrt[3]{(n + 2)^{2}} + \sqrt[3]{(n + 2) (n + 1)} + \sqrt[3]{(n + 1)^{2}})} = \\ = \frac{1}{\sqrt[3]{n^{3} + 4 n^{2} + 2 n} + \sqrt[3]{n^{3} + 3 n^{2} + 2 n} + \sqrt[3]{n^{3} + 2 n^{2} + n}} = \\ = \frac{1}{n} \cdot \frac{1}{\sqrt[3]{1 + \frac{4}{n} + \frac{2}{n^{2}}} + \sqrt[3]{1 + \frac{3}{n} + \frac{2}{n^{2}}} + \sqrt[3]{1 + \frac{2}{n} + \frac{1}{n^{2}}}} \\ lim_{n \to \infty} d_{n} = lim_{n \to \infty} \frac{1}{n} \cdot lim_{n \to \infty} \frac{1}{\underset{\to 1}{\underset{⏟}{\sqrt[3]{1 + \underset{\to 0}{\underset{⏟}{\frac{4}{n}}} + \underset{\to 0}{\underset{⏟}{\frac{2}{n^{2}}}}}}} + \underset{\to 1}{\underset{⏟}{\sqrt[3]{1 + \underset{\to 0}{\underset{⏟}{\frac{3}{n}}} + \underset{\to 0}{\underset{⏟}{\frac{2}{n^{2}}}}}}} + \underset{\to 1}{\underset{⏟}{\sqrt[3]{1 + \underset{\to 0}{\underset{⏟}{\frac{2}{n}}} + \underset{\to 0}{\underset{⏟}{\frac{1}{n^{2}}}}}}}} = \\ = 0 \cdot \frac{1}{3} = 0 \\ lim_{n \to \infty} d_{n} = 0 \end{matrix}

3e

\begin{matrix} p_{n} = \sqrt[n]{2^{n} + 3^{n} + \dots + 10^{n}} \\ Let s_{n} = 2^{n} + 3^{n} + \dots + 10^{n} \\ lim_{n \to \infty} p_{n} = lim_{n \to \infty} \frac{s_{n + 1}}{s_{n}} \\ \frac{2^{n + 1}}{s_{n}} = \frac{2}{1 + \underset{\to \infty}{\underset{⏟}{{(\frac{3}{2})}^{n}}} + \dots + \underset{\to \infty}{\underset{⏟}{{(\frac{10}{2})}^{n}}}} \to 0 \\ \dots \\ \frac{9^{n + 1}}{s_{n}} = \frac{9}{\underset{\to 0}{\underset{⏟}{{(\frac{2}{9})}^{n}}} + \dots + 1 + \underset{\to \infty}{\underset{⏟}{{(\frac{10}{9})}^{n}}}} \to 0 \\ \frac{10^{n + 1}}{s_{n}} = \frac{10}{\underset{\to 0}{\underset{⏟}{{(\frac{2}{10})}^{n}}} + \underset{\to 0}{\underset{⏟}{{(\frac{3}{10})}^{n}}} \dots + 1} \to 10 \\ ⟹ \frac{s_{n + 1}}{s_{n}} = \underset{\to 0}{\underset{⏟}{\frac{2^{n + 1}}{s_{n}}}} + \underset{\to 0}{\underset{⏟}{\frac{3^{n + 1}}{s_{n}}}} + \dots + \underset{\to 10}{\underset{⏟}{\frac{10^{n + 1}}{s_{n}}}} ⟹ lim_{n \to \infty} \frac{s_{n + 1}}{s_{n}} = 10 \\ ⟹ lim_{n \to \infty} p_{n} = 10 \end{matrix}

3f

\begin{matrix} q_{n} = \frac{\sin (\cos (2 n^{2} + 1))}{\sqrt{2 n + 1}} \\ - \frac{1}{\sqrt{2 n + 1}} \leq q_{n} \leq \frac{1}{\sqrt{2 n + 1}} \\ | \frac{- 1}{\sqrt{2 n + 1}} - 0 | = | \frac{1}{\sqrt{2 n + 1}} - 0 | = \frac{1}{\sqrt{2 n + 1}} < \frac{1}{\sqrt{2 n}} \underset{n > N}{<} \frac{1}{\sqrt{2 N}} \\ Let N = \frac{1}{2 ε^{2}}, ε > 0 \\ | \frac{- 1}{\sqrt{2 n + 1}} - 0 | = | \frac{1}{\sqrt{2 n + 1}} - 0 | < \frac{1}{\sqrt{\frac{2}{2 ε^{2}}}} = \frac{1}{\frac{1}{ε}} = ε \\ ⟹ \forall ε > 0 : \exists N : \forall n > N : | \frac{- 1}{\sqrt{2 n + 1}} - 0 | = | \frac{1}{\sqrt{2 n + 1}} - 0 | < ε \\ ⟹ \underset{\to 0}{\underset{⏟}{- \frac{1}{\sqrt{2 n + 1}}}} \leq q_{n} \leq \underset{\to 0}{\underset{⏟}{\frac{1}{\sqrt{2 n + 1}}}} \\ ⟹ lim_{n \to \infty} q_{n} = 0 \end{matrix}

4a

\begin{matrix} Prove by definition: lim_{n \to \infty} \ln (n) = \infty \\ Proof: \\ Definition: \forall M > 0 : \exists N : \forall n > N : a_{n} > M \\ Let N = e^{M}, M > 0 \\ \ln (n) \underset{n > N}{>} \ln (N) = \ln (e^{M}) = M \\ ⟹ \forall M > 0 : \exists N : \forall n > N : \ln (n) > M \\ ⟺ lim_{n \to \infty} \ln (n) = \infty \end{matrix}

4b

\begin{matrix} Given: a_{n} \to \infty, m \in N \\ Prove by definition: lim_{n \to \infty} \sqrt[m]{a_{n}} = \infty \\ Proof: \\ a_{n} \to \infty ⟺ \forall M > 0 : \exists N : \forall n > N : a_{n} > M \\ Let M_{1} = M^{m} \\ a_{n} \underset{n > N}{>} M_{1} ⟹ \sqrt[m]{a_{n}} \underset{n > N}{>} \sqrt[m]{M_{1}} = \sqrt[m]{M^{m}} = M \\ ⟹ \forall M > 0 : \exists N : \forall n > N : \sqrt[m]{a_{n}} > M \\ ⟺ lim_{n \to \infty} \sqrt[m]{a_{n}} = \infty \end{matrix}

4c

\begin{matrix} Given: a_{n} \to \infty, b_{n} \to \infty \\ Prove be definition: lim_{n \to \infty} a_{n} + b_{n} = \infty \\ Proof: \\ a_{n} \to \infty ⟺ \forall M > 0 : \exists N_{1} : \forall n > N_{1} : a_{n} > \frac{M}{2} \\ b_{n} \to \infty ⟺ \forall M > 0 : \exists N_{2} : \forall n > N_{2} : b_{n} > \frac{M}{2} \\ ⟹ \forall M > 0 : \exists N = m a x (N_{1}, N_{2}) : \forall n > N : a_{n} + b_{n} > M \\ ⟺ lim_{n \to \infty} a_{n} + b_{n} = \infty \end{matrix}

a_{n} \to \infty, b_{n} \to - \infty

5a

\begin{matrix} Example of a_{n} + b_{n} \to \infty : \\ a_{n} = 2 n^{2} \to \infty \\ b_{n} = - n^{2} \to - \infty \\ a_{n} + b_{n} = n^{2} \to \infty \end{matrix}

5b

\begin{matrix} Example of a_{n} + b_{n} \to - \infty : \\ a_{n} = n^{2} \to \infty \\ b_{n} = - 2 n^{2} \to - \infty \\ a_{n} + b_{n} = - n^{2} \to - \infty \end{matrix}

5c

\begin{matrix} Example of a_{n} + b_{n} \to 0 : \\ a_{n} = n^{2} \to \infty \\ b_{n} = - n^{2} \to - \infty \\ a_{n} + b_{n} = 0 \to 0 \end{matrix}

5d

\begin{matrix} Example of a_{n} + b_{n} \to 10 : \\ a_{n} = n^{2} + 10 \to \infty \\ b_{n} = - n^{2} \to - \infty \\ a_{n} + b_{n} = 10 \to 10 \end{matrix}

5e

\begin{matrix} Example of ∄ lim_{n \to \infty} a_{n} + b_{n} \\ a_{n} = {\begin{cases} n & n is odd \\ n^{2} & n is even \end{cases} \to \infty \\ b_{n} = {\begin{cases} - n^{2} & n is odd \\ - n & n is even \end{cases} \to - \infty \\ a_{n} + b_{n} = {\begin{cases} \overset{\to - \infty}{\overset{⏞}{n - n^{2}}} & n is odd \\ \underset{\to \infty}{\underset{⏟}{n^{2} - n}} & n is even \end{cases} ⟹ ∄ lim_{n \to \infty} a_{n} + b_{n} \end{matrix}