Infi-1 25

Lemma

\begin{matrix} \forall a, b > 0 : lim_{x \to \infty} \frac{(\ln x)^{a}}{x^{b}} = 0 \\ Proof: \\ lim_{x \to \infty} \frac{(\ln x)^{a}}{x^{b}} = lim_{x \to \infty} {(\frac{\ln (x)}{x^{b / a}})}^{a} \\ t \to 0 ⟹ t^{a} \to 0^{a} = 0 \\ lim_{x \to \infty} \frac{\ln x}{x^{b / a}} \overset{L}{=} lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{b}{a} \cdot x^{\frac{b}{a} - 1}} = lim_{x \to \infty} \frac{1}{\frac{b}{a} \cdot x^{b / a}} = 0 \\ ⟹ lim_{x \to \infty} \frac{(\ln x)^{a}}{x^{b}} = 0 \end{matrix}

\begin{matrix} \forall a > 1, b > 0 : lim_{x \to \infty} \frac{x^{b}}{a^{x}} = 0 \\ Proof: \\ lim_{x \to \infty} \frac{x^{b}}{a^{x}} = lim_{x \to \infty} {(\frac{x}{a^{x / b}})}^{b} \\ lim_{x \to \infty} \frac{x}{a^{x / b}} \overset{L}{=} lim_{x \to \infty} \frac{1}{\ln (a) \cdot a^{x / b} \cdot \frac{1}{b}} = 0 \\ ⟹ lim_{x \to \infty} \frac{x^{b}}{a^{x}} = 0 \end{matrix}

\begin{matrix} lim_{x \to 0} \frac{a^{x} - 1}{x} = \ln a \\ \sum \frac{\ln (n)}{n} \geq \sum \frac{1}{n} ⟹ \sum \frac{\ln (n)}{n} diverges \end{matrix}

\begin{matrix} lim_{n \to \infty} (e^{1 / n} - e^{- 1 / n})^{\sin (1 / n)} \\ Let x_{n} = n \\ Let f (x_{n}) = (e^{1 / n} - e^{- 1 / n})^{\sin (1 / n)} \\ Then lim_{n \to \infty} (e^{1 / n} - e^{- 1 / n})^{\sin (1 / n)} = lim_{x \to \infty} f (x) \\ lim_{x \to \infty} (e^{1 / x} - e^{- 1 / x})^{\sin (1 / x)} = lim_{x \to \infty} e^{\sin (1 / x) \cdot \ln (e^{1 / x} - e^{- 1 / x})} \\ lim_{x \to \infty} \sin (1 / x) \cdot \ln (e^{1 / x} - e^{- 1 / x}) = lim_{x \to \infty} \frac{\ln (e^{1 / x} - e^{- 1 / x})}{{(\sin (\frac{1}{x}))}^{- 1}} \\ Let t = \frac{1}{x}, t \to 0^{+} \\ lim_{t \to 0} \frac{\ln (e^{t} - e^{- t})}{(\sin t)^{- 1}} \overset{L}{=} lim_{t \to 0} \frac{\frac{e^{t} + e^{- t}}{e^{t} - e^{- t}}}{- (\sin t)^{- 2} \cdot \cos t} = - lim_{t \to 0} \frac{(e^{t} + e^{- t}) (\sin t)^{2}}{(e^{t} - e^{- t}) \cos t} = \\ \overset{L}{=} - lim_{t \to 0} \frac{e^{t} + e^{- t}}{\cos t} \cdot lim_{t \to 0} \frac{(\sin t)^{2}}{e^{t} - e^{- t}} = - 2 \cdot lim_{t \to 0} \frac{2 \overset{\to 0}{\overset{⏞}{\sin t}} \cdot \overset{\to 1}{\overset{⏞}{\cos t}}}{\underset{\to 2}{\underset{⏟}{e^{t} + e^{- t}}}} = - 2 \cdot 0 = 0 \\ ⟹ lim_{x \to \infty} e^{\sin (1 / x) \cdot \ln (e^{1 / x} - e^{- 1 / x})} = e^{0} = 1 \\ ⟹ lim_{n \to \infty} (e^{1 / n} - e^{- 1 / n})^{\sin (1 / n)} = 1 \end{matrix}

\begin{matrix} \sum_{n = 1}^{\infty} \frac{s i n (n) \cdot \ln^{2} (n)}{n} \\ \sum_{n = 1}^{\infty} \sin n is bounded \\ \frac{\ln^{2} (n)}{n} \to 0 \\ Let f (x) = \frac{\ln^{2} (x)}{x} \\ f^{'} (x) = \frac{(x \cdot 2 \ln (x) \cdot \frac{1}{x} - \ln^{2} (x))}{x^{2}} = \frac{\ln x}{x^{2}} \cdot (2 - \ln x) \\ x > e^{2} ⟹ 2 - \ln x < 0 ⟹ f^{'} (x) < 0 \\ ⟹ f (x) is monotonically decreasing after e^{2} \\ ⟹ \frac{\ln^{2} (n)}{n} is monotonically decreasing after e^{2} \\ ⟹ By the Dirichlet’s test: \sum_{n = 1}^{\infty} \frac{s i n (n) \cdot \ln^{2} (n)}{n} converges \end{matrix}