Infi-1 10

\begin{matrix} (\begin{matrix} _{\sum a_{n}} \^{\sum | a_{n} |} & Converge & Diverge \\ Converge & Converge abs & Converge cond \\ Diverge & X & D i v e r g e \end{matrix}) \end{matrix}

\begin{matrix} \sum_{n = 1}^{\infty} (- 1)^{n} \cdot \frac{1}{\sqrt{n}} converge or diverge? \\ \sum_{n = 1}^{\infty} | (- 1)^{n} \cdot \frac{1}{\sqrt{n}} | = \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n}} diverge \\ a_{n} = \frac{1}{\sqrt{n}} \to 0 is monotonically decreasing \\ ⟹ By the alternating series test: \\ \sum_{n = 1}^{\infty} (- 1)^{n} \cdot \frac{1}{\sqrt{n}} converge conditionally \end{matrix}

\begin{matrix} \sum_{n = 2}^{\infty} \frac{(- 1)^{n}}{n \ln^{3} n} \\ \sum_{n = 2}^{\infty} | \frac{(- 1)^{n}}{n \ln^{3} n} | = \sum_{n = 2}^{\infty} \frac{1}{n \ln^{3} n} \\ a_{n} = \frac{1}{n \ln^{3} n} ⟹ a_{2^{n}} = \frac{1}{2^{n} \ln^{3} (2^{n})} = \frac{1}{2^{n} n^{3} \ln^{3} 2} \\ ⟹ \sum_{n = 2}^{\infty} \frac{2^{n}}{2^{n} n^{3} \ln^{3} 2} = \frac{1}{\ln^{3} 2} \sum_{n = 2}^{\infty} \frac{1}{n^{3}} converge \\ ⟹ \sum_{n = 2}^{\infty} \frac{(- 1)^{n}}{n \ln^{3} n} converge absolutely \end{matrix}

\begin{matrix} \sum_{n = 1}^{\infty} \frac{\sin (2 n)}{\sqrt{n}} \\ \sum_{n = 1}^{\infty} \sin (2 n) is bounded \\ a_{n} = \frac{1}{\sqrt{n}} \to 0 is monotonically decreasing \\ ⟹ By Dirichlet’s test: \sum_{n = 1}^{\infty} \frac{\sin (2 n)}{\sqrt{n}} converge \end{matrix}

\begin{matrix} \frac{1}{1} + \frac{1}{3} - \frac{2}{5} + \frac{1}{7} + \frac{1}{9} - \frac{2}{11} + \dots \\ \sum_{n = 1}^{\infty} \frac{b_{n}}{2 n - 1} \\ b_{n} = {\begin{cases} 1 & n \mod 3 = 1 \\ 1 & n \mod 3 = 2 \\ - 2 & 3 ∣ n \end{cases} \\ Partial sums of b_{n} : S_{N} = 1, 2, 0, 1, 2, 0, \dots \\ ⟹ | S_{N} | \leq 2 \\ a_{n} = \frac{1}{2 n - 1} \to 0 is monotonically decreasing \\ ⟹ By Dirichlet’s test: \sum_{n = 1}^{\infty} \frac{b_{n}}{2 n - 1} \end{matrix}

\begin{matrix} lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = \frac{1}{2} \\ Proof: \\ lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = lim_{x \to 0} \frac{1 - \cos^{2} x}{x (1 + \cos x)} = lim_{x \to 0} {(\frac{\sin x}{x})}^{2} \frac{1}{1 + \cos x} = 1^{2} \cdot \frac{1}{2} = \frac{1}{2} \end{matrix}

\begin{matrix} lim_{x \to 0} (\frac{1}{\sin x} - \frac{1}{\tan x}) = lim_{x \to 0} (\frac{1}{\sin x} - \frac{\cos x}{\sin x}) = lim_{x \to 0} \frac{1 - \cos x}{\sin x} = \\ = lim_{x \to 0} \underset{\to 1}{\underset{⏟}{\frac{x}{\sin x}}} \underset{\to \frac{1}{2}}{\underset{⏟}{\frac{1 - \cos x}{x^{2}}}} \underset{\to 0}{\underset{⏟}{x}} = 1 \cdot \frac{1}{2} \cdot 0 = 0 \end{matrix}

\begin{matrix} lim_{x \to 2} \frac{\sin (6 x - 12)}{\sqrt{x} - \sqrt{2}} = \frac{0}{2} = lim_{x \to 2} \underset{\to 1}{\underset{⏟}{\frac{\sin (6 x - 12)}{6 x - 12}}} \cdot \frac{6 x - 12}{\sqrt{x} - \sqrt{2}} = \\ = lim_{x \to 2} 6 \cdot \frac{x - 2}{x - 2} \cdot (\sqrt{x} + \sqrt{2}) = 12 \sqrt{2} \end{matrix}

\begin{matrix} lim_{x \to \infty} \frac{x \cdot 2^{- x} + \arctan x}{x} = \\ = lim_{x \to \infty} \underset{\to 0}{\underset{⏟}{2^{- x}}} + \underset{\to 0}{\underset{⏟}{\frac{\overset{- π / 2 \leq \arctan x \leq π / 2}{\overset{⏞}{\arctan x}}}{x}}} = 0 \end{matrix}