Infi-2 8

\begin{matrix} \sum_{n = 0}^{\infty} \frac{2^{n} x^{n}}{n!} \\ lim_{n \to \infty} \frac{\frac{2^{n + 1} x^{n + 1}}{(n + 1)!}}{\frac{2^{n} x^{n}}{n!}} = lim_{n \to \infty} \frac{2 x}{n + 1} = 0 \\ \forall x \in R : \sum_{n = 0}^{\infty} \frac{2^{n} x^{n}}{n!} \to S (x) \end{matrix}

\begin{matrix} \sum_{n = 1}^{\infty} \frac{\sin (n! x)}{n^{3} + n + 1}, x \in [- 2 π, 2] \\ | \frac{\sin (n! x)}{n^{3} + n + 1} | \leq \frac{1}{n^{3} + n + 1} \leq \frac{1}{n^{3}} \\ \sum_{n = 1}^{\infty} \frac{1}{n} \to M ⟹ \sum_{n = 1}^{\infty} \frac{\sin (n! x)}{n^{3} + n + 1} ⇉ S (x) \end{matrix}

\begin{matrix} \sum_{n = 2}^{\infty} \frac{x^{4} + x^{2}}{n \ln^{2} (n)}, x \in [- 7, 2] \\ | \frac{x^{4} + x^{2}}{n \ln^{2} (n)} | \leq \frac{7^{4} + 7^{2}}{n \ln^{2} (n)} \\ \sum_{n = 2}^{\infty} \frac{1}{n \ln^{2} (n)} \to M by the Cauchy’s condensation test \\ ⟹ \sum_{n = 2}^{\infty} \frac{x^{4} + x^{2}}{n \ln^{2} (n)} ⇉ S (x) \end{matrix}

\begin{matrix} \sum_{n = 0}^{\infty} \frac{\sin (x)}{(1 + x)^{n}} \\ \sum_{n = 0}^{\infty} \frac{\sin (x)}{(1 + x)^{n}} = \sin (x) \cdot \sum_{n = 0}^{\infty} t^{n} = \frac{1}{1 - \frac{1}{1 + x}} = \sin (x) \cdot \frac{1 + x}{x} \\ Let f (x) = {\begin{matrix} \sin (x) \cdot \frac{1 + x}{x} & x \neq 0 \\ 0 & x = 0 \end{matrix} \\ f is not continuous ⟹ x \in [0, π) ⟹ \sum_{n = 0}^{\infty} \frac{\sin (x)}{(1 + x)^{n}} ⇉̸ f (x) \\ Let d_{N} = sup_{x \in (0, π)} | S_{N} (x) - f (x) | \\ d_{N}^{'} = sup_{x \in [0, π)} | S_{N} (x) - f (x) | = max {d_{N}, | S_{N} (0) - f (0) |} = d_{N} \\ d_{n}^{'} ↛ 0 ⟹ d_{n} ↛ 0 ⟹ x \in (0, π) ⟹ \sum_{n = 0}^{\infty} \frac{\sin (x)}{(1 + x)^{n}} ⇉̸ f (x) \end{matrix}

\begin{matrix} \sum_{n = 1}^{\infty} \frac{1}{n \cdot 3^{n}} \\ Let x = \frac{1}{3} \\ \sum_{n = 1}^{\infty} \frac{1}{n} \cdot x^{n} = \sum_{n = 1}^{\infty} \int_{0}^{x} t^{n - 1} d t \underset{\sum_{n = 1}^{\infty} t^{n - 1} ⇉ S (x)}{\overset{t \in [- \frac{1}{2}, \frac{1}{2}]}{=}} \int_{0}^{x} \sum_{n = 1}^{\infty} t^{n - 1} d t = \int_{0}^{x} \frac{1}{1 - t} d t = - \ln | 1 - x | \\ ⟹ \sum_{n = 1}^{\infty} \frac{1}{n \cdot 3^{n}} = \ln (\frac{3}{2}) \end{matrix}

\begin{matrix} \sum_{n = 1}^{\infty} n x^{n} \\ \sum_{n = 1}^{\infty} n x^{n} = x \cdot \sum_{n = 1}^{\infty} n x^{n - 1} = \frac{x}{(1 - x)^{2}} \end{matrix}