Infi-1 16

\begin{matrix} S_{N} = \sum_{n = 1}^{N} \sin (n) = \sin (1) + \sin (2) + \dots + \sin (N) \\ 2 \sin (x) \sin (y) = \cos (x - y) - \cos (x + y) \\ 2 \sin (1) S_{N} = 2 \sin (1) \sin (1) + 2 \sin (1) \sin (2) + \dots + 2 \sin (1) \sin (N) = \\ = \cos (0) - \cos (2) + \cos (1) - \cos (3) + \cos (2) - \cos (4) + \dots + \cos (N - 1) - \cos (N + 1) = \\ = \cos (0) + \cos (1) - \cos (N) - \cos (N + 1) \\ ⟹ S_{N} = \frac{\cos (0) + \cos (1) - \cos (N) - \cos (N + 1)}{2 \sin (1)} \\ - 2 \leq - \cos (N) - \cos (N + 1) \leq 2 ⟹ \frac{\cos (0) + \cos (1) - 2}{2 \sin (1)} \leq S_{N} \leq \frac{\cos (0) + \cos (1) + 2}{2 \sin (1)} \end{matrix}

Alternating series test (Leibniz criterion) #theorem

\begin{matrix} \sum (- 1)^{n} \\ a_{n} is monotonically non-increasing and a_{n} \to 0 \\ \sum (- 1)^{n} a_{n} \to M_{1} \end{matrix}

\begin{matrix} r_{N} = \sum a_{n} - S_{N} = \sum_{n = N + 1}^{\infty} a_{n} \\ S_{N} \to \sum a_{n} ⟹ r_{N} \to 0 \end{matrix}

\begin{matrix} \sum \frac{(- 1)^{n + 1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \\ S_{4} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \\ r_{4} = \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \dots \leq \frac{1}{5} \\ | r_{N} | \leq | a_{N + 1} | \end{matrix}

\begin{matrix} \ln (2) = \sum \frac{(- 1)^{n + 1}}{n} \end{matrix}

\begin{matrix} \sum (- 1)^{n} = - 1 + 1 - 1 + 1 - 1 + \dots = \\ = (- 1 + 1) + (- 1 + 1) + \dots = 0 = \\ = - 1 + (1 - 1) + (1 - 1) + (1 - 1) + \dots = - 1 = \\ = 1 + (- 1 + 1) + (- 1 + 1) + (- 1 + 1) + \dots = 1 \\ 0 = - 1 = 1 ? ? ? \\ ⟹ In non-convergent series grouping or changing order is not allowed \end{matrix}

\begin{matrix} \sum \frac{(- 1)^{n + 1}}{n} = a \\ a = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \\ (1 - \frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{6} - \frac{1}{8}) + \dots + (\frac{1}{2 n + 1} - \frac{1}{4 n + 2} - \frac{1}{4 n + 4}) = \\ = (\frac{1}{2} - \frac{1}{4}) + \dots + (\frac{1}{4 n + 2} - \frac{1}{4 n + 4}) = \frac{1}{2} (1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots) = \frac{1}{2} a \\ ⟹ In convergent series changing order is not allowed \end{matrix}

\begin{matrix} \sum | a_{n} | \to M_{1} \\ It is allowed to change the order of terms in absolutely convergent series \end{matrix}

\begin{matrix} If series converge conditionally, then for all w \in R exists an order of terms such that \\ \sum a_{n}^{'} \to w \end{matrix}