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\begin{matrix} a_{n} \to \infty \\ b_{n} is bottom-limited/bottom-bounded (b_{n} \geq B) \\ Prove by definition: a_{n} + b_{n} \to \infty \\ Proof: \\ a_{n} \to \infty ⟹ \forall M_{1} \in R : \exists N \in N : \forall n > N : a_{n} > M_{1} \\ b_{n} is bottom-bounded/lower-bounded/bottomo-limited/etc. \\ ⟹ \exists B \in R : \forall n \in N : b_{n} \geq B \\ Let B \in R : b_{n} \geq B \\ ⟹ \forall n \in N : a_{n} + b_{n} \geq a_{n} + B \\ Let M \in R \\ Let M_{1} = M - B \\ \forall M_{1} : \exists N : \forall n > N : a_{n} > M_{1} \\ ⟹ a_{n} > M - B ⟹ a_{n} + b_{n} \geq a_{n} + B > M - B + B = M \\ ⟹ \forall M : \exists N : \forall n > N : a_{n} + b_{n} > M \\ ⟹ lim_{n \to \infty} a_{n} + b_{n} = \infty \end{matrix}

\begin{matrix} a_{n} = {\begin{cases} a_{1} > 1 \\ a_{n + 1} = 2 a_{n} - 1 \end{cases} \\ Prove: a_{n} is monotonically strictly increasing \\ Find lim_{n \to \infty} a_{n} \\ Solution: \\ Base case. a_{1} > 1 \\ Induction step. Let a_{n} > 1 \\ ⟹ a_{n + 1} = 2 a_{n} - 1 > 2 - 1 = 1 ⟹ a_{n + 1} > 1 \\ ⟹ By induction: \forall n \in N : a_{n} > 1 \\ a_{n + 1} - a_{n} = 2 a_{n} - 1 - a_{n} = a_{n} - 1 > 0 \\ ⟹ a_{n + 1} > a_{n} ⟹ a_{n} is monotonically strictly increasing \\ Let lim_{n \to \infty} a_{n} = L \in R \\ ⟹ lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} a_{n} = L \\ ⟹ L = lim_{n \to \infty} 2 a_{n} - 1 = 2 L - 1 \\ L = 2 L - 1 ⟹ L = 1 \\ a_{1} > 1 ⟹ a_{1} = 1 + α, α > 0 \\ \forall n \in N : a_{n + 1} > a_{1} ⟹ a_{n + 1} > 1 + α ⟹ lim_{n \to \infty} a_{n + 1} \geq 1 + α > 1 \\ ⟹ lim_{n \to \infty} a_{n} \neq 1 ⟹ L \neq 1 ⟹ L \notin R \\ a_{n} is monotonically strictly increasing and a_{n} > 0 \\ ⟹ a_{n} converges in the broadest sense ⟹ lim_{n \to \infty} a_{n} = \infty \end{matrix}

\begin{matrix} Find lim_{n \to \infty} \frac{\sqrt{(n + 1)^{3}} - \sqrt{n^{3}}}{\sqrt{n + 1}} \\ Solution: \\ a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) \\ lim_{n \to \infty} \frac{\sqrt{(n + 1)^{3}} - \sqrt{n^{3}}}{\sqrt{n + 1}} = lim_{n \to \infty} \frac{(\sqrt{(n + 1)} - \sqrt{n}) (\sqrt{(n + 1)^{2}} + \sqrt{n (n + 1)} + \sqrt{n^{2}})}{\sqrt{n + 1}} = \\ = lim_{n \to \infty} \frac{(\sqrt{(n + 1)} - \sqrt{n}) (\sqrt{(n + 1)} + \sqrt{n}) (n + 1 + \sqrt{n (n + 1)} + n)}{\sqrt{n + 1} (\sqrt{(n + 1)} + \sqrt{n})} = \\ = lim_{n \to \infty} \frac{(n + 1 - n) (n + 1 + \sqrt{n (n + 1)} + n)}{n + 1 + \sqrt{n (n + 1)}} = lim_{n \to \infty} \frac{2 n + 1 + \sqrt{n^{2} + n}}{n + 1 + \sqrt{n^{2} + n}} = \\ = lim_{n \to \infty} \frac{2 + \overset{\to 0}{\overset{⏞}{\frac{1}{n}}} + \overset{\to 1}{\overset{⏞}{\sqrt{1 + \frac{1}{n}}}}}{1 + \underset{\to 0}{\underset{⏟}{\frac{1}{n}}} + \underset{\to 1}{\underset{⏟}{\sqrt{1 + \frac{1}{n}}}}} = \frac{2 + 1}{1 + 1} = \frac{3}{2} \\ ⟹ lim_{n \to \infty} \frac{\sqrt{(n + 1)^{3}} - \sqrt{n^{3}}}{\sqrt{n + 1}} = \frac{3}{2} \end{matrix}

\begin{matrix} a_{n} is monotonically non-increasing \\ a_{n} > 0 \\ b_{n} = \frac{a_{2 n}}{a_{2 n - 1}} (a_{2 n - 1} - a_{2 n}) \\ Prove that b_{n} converges and find lim_{n \to \infty} b_{n} \\ Solution: \\ Let c_{n} = \frac{a_{2 n}}{a_{2 n - 1}} \\ 0 < a_{2 n} \leq a_{2 n - 1} ⟹ 0 < \frac{a_{2 n}}{a_{2 n - 1}} \leq 1 ⟹ 0 < c_{n} \leq 1 ⟹ | c_{n} | \leq 1 \\ Let d_{n} = a_{2 n - 1} - a_{2 n} \\ a_{n} is monotonically non-increasing and bottom-bounded \\ ⟹ \exists lim_{n \to \infty} a_{n} = L \in R and L \geq 0 \\ \exists lim_{n \to \infty} a_{n} ⟹ lim_{n \to \infty} a_{2 n} = lim_{n \to \infty} a_{2 n - 1} = L \\ ⟹ lim_{n \to \infty} d_{n} = lim_{n \to \infty} a_{2 n - 1} - a_{2 n} = lim_{n \to \infty} a_{2 n - 1} - lim_{n \to \infty} a_{2 n} = L - L = 0 \\ lim_{n \to \infty} b_{n} = lim_{n \to \infty} c_{n} \cdot d_{n} \\ {\begin{matrix} c_{n} is both top- and bottom-bounded \\ d_{n} \to 0 \end{matrix} ⟹ lim_{n \to \infty} c_{n} \cdot d_{n} = 0 \\ ⟹ lim_{n \to \infty} b_{n} = 0 \end{matrix}

\begin{matrix} \sum_{n = 1}^{\infty} a_{n} converges \\ \sum_{n = 1}^{\infty} a_{n} > 0 \\ a_{n} = S_{n}^{2} + S_{n} - 6 \\ Find \sum_{n = 1}^{\infty} a_{n} \\ Solution: \\ S_{n + 1} - S_{n} = a_{n + 1} \\ ⟹ S_{n + 1} - S_{n} = S_{n + 1}^{2} + S_{n + 1} - 6 \\ ⟹ S_{n} = 6 - S_{n + 1}^{2} \\ \sum_{n = 1}^{\infty} a_{n} converges ⟹ \exists lim_{n \to \infty} S_{n} = L \in R \\ lim_{n \to \infty} S_{n + 1} = lim_{n \to \infty} S_{n} = L \\ ⟹ lim_{n \to \infty} S_{n} = lim_{n \to \infty} (6 - \underset{\to L^{2}}{\underset{⏟}{S_{n + 1}^{2}}}) \\ ⟹ L = 6 - L^{2} \\ L^{2} + L - 6 = 0 ⟹ (L + 3) (L - 2) = 0 ⟹ [\begin{matrix} L = - 3 \\ L = 2 \end{matrix} \\ \sum_{n = 1}^{\infty} a_{n} > 0 ⟹ lim_{n \to \infty} S_{n} = L > 0 \\ ⟹ L = 2 ⟹ lim_{n \to \infty} S_{n} = 2 ⟹ \sum_{n = 1}^{\infty} a_{n} = 2 \end{matrix}