Infi-1 5

Infinite sequence limit (sequence diverges)

\begin{matrix} lim_{n \to \infty} a_{n} = \infty ⟺ \forall M \exists N : \forall n > N : a_{n} > M \\ lim_{n \to \infty} a_{n} = - \infty ⟺ lim_{n \to \infty} - a_{n} = \infty \\ ⟺ \forall m \exists N : \forall n > N : a_{n} < m \end{matrix}

\begin{matrix} Prove: lim_{n \to \infty} \sqrt{\frac{n}{26}} = \infty \\ Proof: \\ Let M \\ \sqrt{\frac{n}{26}} \underset{n > N}{>} \sqrt{\frac{N}{26}} \\ Let N = 26 M^{2} \\ \sqrt{\frac{n}{26}} > \sqrt{\frac{26 M^{2}}{26}} = \sqrt{M^{2}} = M \\ ⟹ \forall M : \exists N = 26 M^{2} : \forall n > N, n \in N : \sqrt{\frac{n}{26}} > M \\ ⟺ lim_{n \to \infty} \sqrt{\frac{n}{26}} = \infty \end{matrix}

\begin{matrix} Prove or disprove: lim_{n \to \infty} a_{n} = 0 ⟹ lim_{n \to \infty} \frac{1}{a_{n}} = \pm \infty \\ Disproof: \\ lim_{n \to \infty} (- 1)^{n} \frac{1}{n} = 0 \\ a_{n} = (- 1)^{n} \frac{1}{n} \\ lim_{n \to \infty} \frac{1}{a_{n}} = lim_{n \to \infty} \frac{1}{(- 1)^{n} \frac{1}{n}} = lim_{n \to \infty} (- 1)^{n} \cdot n \\ ∄ lim_{n \to \infty} (- 1)^{n} \cdot n ⟹ ∄ lim_{n \to \infty} \frac{1}{a_{n}} \end{matrix}

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

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

\begin{matrix} \sqrt[3]{n + 1} - 1 \\ a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) \\ ⟹ \sqrt[3]{n + 1} - 1 = \frac{(\sqrt[3]{n + 1} - 1) (\sqrt[3]{(n + 1)^{2}} + \sqrt{n + 1} + 1)}{\sqrt[3]{(n + 1)^{2}} + \sqrt{n + 1} + 1} = \\ = \frac{n}{\sqrt[3]{(n + 1)^{2}} + \sqrt{n + 1} + 1} \end{matrix}

\begin{matrix} \exists N : \forall n > N, n \in N : a_{n} \leq b_{n} \leq c_{n} \\ a_{n} \to L, c_{n} \to L ⟹ b_{n} \to L \end{matrix}

\begin{matrix} lim_{n \to \infty} \frac{\sin^{2} (n^{2})}{n^{3} + 1} = 0 \\ \underset{\to 0}{\underset{⏟}{- \frac{1}{n^{3} + 1}}} \leq \frac{\sin^{2} (n^{2})}{n^{3} + 1} \leq \underset{\to 0}{\underset{⏟}{\frac{1}{n^{3} + 1}}} \\ ⟹ \frac{\sin^{2} (n^{2})}{n^{3} + 1} \to 0 \end{matrix}

\begin{matrix} lim_{n \to \infty} \sqrt[n]{3^{n} + 2^{n} + 4^{n}} \\ \sqrt[n]{4^{n}} \leq \sqrt[n]{3^{n} + 2^{n} + 4^{n}} \leq \sqrt[n]{4^{n} + 4^{n} + 4^{n}} \\ \underset{\to 4}{\underset{⏟}{4}} \leq \sqrt[n]{3^{n} + 2^{n} + 4^{n}} \leq \underset{\to 1}{\underset{⏟}{\sqrt[n]{3}}} \cdot \underset{\to 4}{\underset{⏟}{4}} \\ ⟹ lim_{n \to \infty} \sqrt[n]{3^{n} + 2^{n} + 4^{n}} = 4 \end{matrix}

\begin{matrix} lim_{n \to \infty} (\frac{1}{\sqrt{n^{2} + 1}} + \frac{1}{\sqrt{n^{2} + 2}} + \dots + \frac{1}{\sqrt{n^{2} + n}}) \\ \frac{n}{\sqrt{n^{2} + n}} \leq a_{n} \leq \frac{n}{\sqrt{n^{2} + 1}} \\ \frac{n}{n \sqrt{1 + \frac{1}{n}}} \leq a_{n} \leq \frac{n}{n \sqrt{1 + \frac{1}{n^{2}}}} \\ \underset{\to \frac{1}{1} = 1}{\underset{⏟}{\frac{1}{\underset{\to 1}{\underset{⏟}{\sqrt{1 + \frac{1}{n}}}}}}} \leq a_{n} \leq \underset{\to \frac{1}{1} = 1}{\underset{⏟}{\frac{1}{\underset{\to 1}{\underset{⏟}{\sqrt{1 + \frac{1}{n^{2}}}}}}}} \\ ⟹ lim_{n \to \infty} (\frac{1}{\sqrt{n^{2} + 1}} + \frac{1}{\sqrt{n^{2} + 2}} + \dots + \frac{1}{\sqrt{n^{2} + n}}) = 1 \end{matrix}

\begin{matrix} \forall n \in N : a_{n + 1} - a_{n} \geq 0 - Monotonic non-descending \\ \forall n \in N : \frac{a_{n + 1}}{a_{n}} \geq 1 - Monotonic non-descending \\ \forall n \in N : a_{n + 1} - a_{n} \leq 0 - Monotonic non-ascending \\ \forall n \in N : 0 < \frac{a_{n + 1}}{a_{n}} \leq 1 - Monotonic non-descending \end{matrix}

\begin{matrix} a_{n} = \frac{1}{n} + \frac{1}{n + 1} + \dots + \frac{1}{3 n} \\ Prove: lim_{n \to \infty} a_{n} = L \\ 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 descending \\ \frac{2}{3} \leq \frac{2 n + 1}{3 n} \leq \frac{1}{n} + \frac{1}{n + 1} + \dots + \frac{1}{3 n} \\ ⟹ a_{n} is lower-bounded ⟹ lim_{n \to \infty} a_{n} = L \geq \frac{2}{3} \end{matrix}

\begin{matrix} n \cdot \frac{n}{n^{2} + n} = \frac{n^{2}}{n^{2} + n} = \frac{1}{1 + \frac{1}{n}} \end{matrix}