Infi-1 12

Partial sum #definition

\begin{matrix} S_{N} = \sum_{n = 1}^{N} a_{n} \\ \sum_{n = 1}^{\infty} = lim_{N \to \infty} S_{N} \end{matrix}

\begin{matrix} \sum_{n = 1}^{N} C a_{n} = C \sum_{n = 1}^{N} a_{n} \end{matrix}

\begin{matrix} \sum_{n = 1}^{N} a_{n} + \sum_{n = 1}^{N} b_{n} = \sum_{n = 1}^{N} (a_{n} + b_{n}) = C_{N} \\ A_{N} = \sum_{n = 1}^{N} a_{n} \\ B_{N} = \sum_{n = 1}^{N} b_{n} \\ A_{N} \to L, B_{N} \to M \\ C_{N} = \sum_{n = 1}^{N} (a_{n} + b_{n}) = a_{1} + b_{1} + \dots + a_{N} + b_{N} = \\ = \underset{\sum_{n = 1}^{N} a_{n}}{\underset{⏟}{a_{1} + a_{2} + \dots + a_{N}}} + \underset{\sum_{n = 1}^{N} b_{n}}{\underset{⏟}{b_{1} + b_{2} + \dots + b_{N}}} = A_{N} + B_{N} \to L + M \end{matrix}

\begin{matrix} \sum_{n = 1}^{\infty} a_{n} \to L ⟹ a_{n} \to 0 \\ S_{N} = \sum_{n = 1}^{N} a_{n}, S_{N} \to L \\ S_{N} - S_{N - 1} = a_{N} \\ S_{N} - S_{N - 1} \to L - L = 0 ⟹ a_{N} \to 0 \\ a_{N} ↛ 0 ⟹ \sum_{n = 1}^{\infty} a_{n} ↛ L \end{matrix}

{(1 + \frac{a}{n})}^{n} \to e^{a}

\sum_{n = 1}^{\infty} \frac{1}{n^{a}} \to L ⟺ a > 1

\begin{matrix} \sum_{n = 1}^{\infty} a_{n} \to L ⟺ S_{N} is a Cauchy’s sequence \\ (\forall ε > 0 : \exists N_{ε} : \forall N, M > N_{ε} : | S_{N} - S_{M} | < ε) \\ If M > N : \\ S_{M} = S_{N} + \sum_{n = N + 1}^{M} a_{n} \\ | S_{N} - S_{M} | < ε \\ | \sum_{n = N + 1}^{M} a_{n} | < ε \end{matrix}

\begin{matrix} \sum a_{n}, \sum b_{n} : \\ 0 \leq a_{n} \leq b_{n} \\ \sum b_{n} \to L ⟹ \sum a_{n} \to M \\ \sum a_{n} ↛ M ⟹ \sum b_{n} ↛ L \\ Proof: \\ 0 \leq a_{n} ⟹ A_{N} - A_{N - 1} = a_{N} \geq 0 \\ A_{N} > A_{N} ⟹ A_{N} is monotonically non-descending \\ \sum_{n = 1}^{N} b_{n} = B_{N} \\ \sum b_{n} \to L ⟹ \exists C : B_{N} \leq C \\ A_{N} \leq B_{N} \leq C ⟹ A_{N} < C \\ ⟹ A_{N} is monotonically non-descending and upper-boundeed ⟹ A_{N} \to M \end{matrix}

\begin{matrix} \sum a_{n}, \sum b_{n} \\ 0 \leq a_{n}, b_{n} \\ \exists lim_{n \to \infty} \frac{a_{n}}{b_{n}} = L \\ L > 0 ⟹ (\sum a_{n} \to M_{1} ⟺ \sum b_{n} \to M_{2}) \\ L = 0 ⟹ (\sum a_{n} \to M_{1} ⟸ \sum b_{n} \to M_{2}) \\ L = \infty ⟹ (\sum a_{n} \to M_{1} ⟹ \sum b_{n} \to M_{2}) \end{matrix}