Infi-1 10

Cauchy's sequence #definition

\begin{matrix} \forall ε > 0 \exists N_{ε} : \forall n, m > N_{ε} : | a_{n} - a_{m} | < ε \end{matrix}

Cauchy's criterion #theorem

\begin{matrix} {a_{n}} \subseteq R \\ a_{n} converges ⟺ a_{n} is a Cauchy’s sequence \\ Proof: \\ Let a_{n} \to L \\ \forall ε > 0 \exists N_{ε} : \forall n > N_{ε} : | a_{n} - L | < \frac{ε}{2} \\ Let m \geq n \\ | a_{n} - a_{m} | = | a_{n} - L - a_{m} + L | \leq \underset{< \frac{ε}{2}}{\underset{⏟}{| a_{n} - L |}} + \underset{= | a_{m} - L | < \frac{ε}{2}}{\underset{⏟}{| L - a_{m} |}} \\ ⟹ | a_{n} - a_{m} | < ε ⟺ a_{n} is a Cauchy’s sequence \end{matrix}

\begin{matrix} Let a_{n} be a Cauchy’s sequence \\ 1. & Cauchy’s sequence is bounded \\ Proof: \\ By definition of the limit: \exists ε = 7 : \exists N_{ε} : \forall n, m > N_{ε} : | a_{n} - a_{m} | < 7 \\ Let z = a_{⌈ N_{ε} ⌉ + 1} \\ Then - 7 < a_{n} - z < 7 ⟺ \forall n > N_{ε} : z - 7 \leq a_{n} \leq z + 7 \\ A = {a_{1}, a_{2}, a_{3}, \dots, a_{⌈ N_{ε} ⌉}} is a finite set ⟹ \exists m = m i n (A), M = m a x (A) \\ ⟹ \forall n \in N : m i n (m, z - 7) \leq a_{n} \leq m a x (M, z + 7) - sequence is bounded \\ 2. & If sequence is bounded, then there exists a convergent subsequence a_{k_{n}} \\ \exists a_{k_{n}} \to L_{k} \\ 3. & Limit of a subsequence is a limit of the whole sequence, if such exists \\ \forall ε > 0 \exists N_{a} : \forall n, m > N_{a} : | a_{n} - a_{m} | < \frac{ε}{2} \\ \forall ε > 0 \exists N_{k} : \forall n > N_{k} : | a_{k_{n}} - L_{k} | < \frac{ε}{2} \\ Let n > N_{ε} = m a x (N_{a}, N_{k}) \\ | a_{n} - L_{k} | = | a_{n} - a_{k_{n}} - L_{k} + a_{k_{n}} | \leq \underset{< \frac{ε}{2}}{\underset{⏟}{| a_{n} - a_{k_{n}} |}} + \underset{< \frac{ε}{2}}{\underset{⏟}{| a_{k_{n}} - L_{k} |}} < ε \\ ⟺ a_{n} \to L_{k} \\ 4. & Proof for 3. via lim^{―} a_{n} \\ Let lim^{―} a_{n} = T > L_{k} \\ Then \exists a_{m_{n}} \to T \\ \exists N_{1} : \forall n > N_{1} : | a_{k_{n}} - L_{k} | < \frac{T - L}{4} \\ \exists N_{2} : \forall n > N_{2} : | a_{m_{n}} - T | < \frac{T - L}{4} \\ \exists N_{3} : \forall n, m > N_{3} : | a_{n} - a_{m} | < \frac{T - L}{4} \\ \exists N_{ε} = m a x (N_{1}, N_{2}, N_{3}) : \forall n, m > N_{ε} : \\ | a_{k_{n}} - L_{k} | < \frac{T - L}{4}, | a_{m_{n}} - T | < \frac{T - L}{4}, | a_{n} - a_{m} | < \frac{T - L}{4} \\ L - \frac{T - L}{4} < a_{k_{n}} < L + \frac{T - L}{4} \\ T - \frac{T - L}{4} < a_{m_{n}} < T + \frac{T - L}{4} \\ \frac{5 L - T}{4} < a_{k_{n}} < \frac{T + 3 L}{4} \\ \frac{3 T + L}{4} < a_{m_{n}} < \frac{5 T - L}{4} \\ a_{m_{n}} - a_{k_{n}} > \frac{3 T + L}{4} - a_{k_{n}} > \frac{3 T + L}{4} - \frac{T + 3 L}{4} = \frac{T - L}{2} > \frac{T - L}{4} \end{matrix}

Series (טורים) #definition

\begin{matrix} Series are infinite sums, corresponding to sequences \\ For example: \\ a_{n} = \frac{1}{2^{n}} \\ S_{n} = 9 \sum_{n = 1}^{\infty} \frac{1}{2^{n}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots \end{matrix}