Infi-1 9

Subsequence #definition

\begin{matrix} Let a_{n} \\ Let a_{k_{n}} : \forall n \in N : k_{n} < k_{n + 1} \\ For example: \\ a_{n} = {\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots} \\ a_{2 n} = {\frac{1}{4}, \frac{1}{16}, \frac{1}{64}, \dots}, k_{n} = 2 n : 2 < 4 < 6 < \dots \\ a_{n + 7} = {\frac{1}{256}, \frac{1}{512}, \dots} \\ k_{n} = n + 7 : 8 < 9 < 10 < \dots \end{matrix}

\begin{matrix} For any sequence there exists a monotonic subsequence \\ Proof: \\ Let x_{m} be a "peak" if \forall n > m : x_{n} \leq x_{m} \\ 1. There are infinitely many "peaks" \\ ⟹ x_{m_{n}} = {x_{m_{1}}, x_{m_{2}}, x_{m_{3}}, \dots}, m_{1} < m_{2} < \dots \\ \forall m_{a}, m_{b} : m_{a} < m_{b} ⟹ x_{m_{a}} \geq x_{m_{b}} \\ ⟹ {x_{m_{1}}, x_{m_{2}}, \dots} - monotonically non-asscending \\ 2. There is a finite number of "peaks" \\ Let x_{n_{1}} be past all of them \\ x_{n_{1}} is not a peak, then \exists n_{2} > n_{1} : x_{n_{2}} > x_{n_{1}} \\ \forall a > n_{2} : x_{n_{a}} is not a "peak" ⟹ \exists n_{b} : x_{n_{a}} > x_{n_{b}} \\ By induction, sequence {x_{n_{1}}, x_{n_{2}}, \dots} - monotonically ascending \end{matrix}

\begin{matrix} Every bounded sequence has a convergent subsequence \\ Proof: \\ \exists a_{k_{n}} : a_{k_{n}} monotonic and bounded \\ ⟹ a_{k_{n}} converges \end{matrix}

\begin{matrix} Limit of any subsequence is equal to the limit of a sequence, if such exists \\ Proof: \\ \dots \end{matrix}

\begin{matrix} If a_{2 n}, a_{2 n - 1} converge to L, then a_{n} converges to L \\ Proof: \\ \forall ε > 0 : \exists N_{2 n} : \forall 2 n > N_{2 n} : | a_{2 n} - L | < ε \\ \forall ε > 0 : \exists N_{2 n - 1} : \forall 2 n - 1 > N_{2 n - 1} : | a_{2 n - 1} - L | < ε \\ ⟹ \forall ε > 0 : \exists N = m a x (N_{2 n}, N_{2 n - 1}) : \forall n > N : | a_{n} - L | < ε \\ ⟺ lim_{n \to \infty} a_{n} = L \end{matrix}

\begin{matrix} Let a_{n}, c \in R \\ C is called a partial limit of a_{n} if there exists a subsequence a_{k_{n}} converging to C \end{matrix}

\begin{matrix} a_{n} = (- 1)^{n} \\ Set of partial limits of this sequence is {1, - 1} \end{matrix}

\begin{matrix} a_{n} = {1, \frac{1}{2}, 1, \frac{1}{2}, \frac{1}{3}, \dots} \\ Set of partial limits of this sequence is {\frac{1}{n} ∣ n \in N} \cup {0} \end{matrix}

\begin{matrix} Supremum of the set of partial limits of a_{n} is called limit superior - l i m s u p (a_{n}) or lim^{―} a_{n} \end{matrix}

\begin{matrix} Infimum of the set of partial limits of a_{n} is called limit superior - l i m i n f (a_{n}) or lim_{―} a_{n} \end{matrix}

\begin{matrix} \overset{―}{l i m} a_{n} = \underset{―}{l i m} a_{n} = L ⟹ lim_{n \to \infty} a_{n} = L \\ Note: also true for converging in the broadest sense \\ Proof: \\ \dots \end{matrix}