Infi-2 15

Uniform convergence of function series

\begin{matrix} S_{N} ⇉ S \\ ⟺ d_{N} = sup_{x \in A} | S_{N} (x) - S (x) | = sup_{x \in A} | \sum_{n = 1}^{N} f_{n} (x) - \sum_{n = 1}^{\infty} f_{n} (x) | = \\ = sup_{x \in A} | \sum_{n = N + 1}^{\infty} f_{n} (x) | = sup_{x \in A} | r_{N} | \end{matrix}

\begin{matrix} \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x}, x \in (- 1, 1) \\ d_{N} = sup_{x \in (- 1, 1)} | \sum_{n = N + 1}^{\infty} x^{n} | \\ \sum_{n = N + 1}^{\infty} x^{n} = x^{N + 1} + x^{N + 2} + \dots = x^{N + 1} \cdot \sum_{n = 0}^{\infty} x^{n} = \frac{x^{N + 1}}{1 - x} \\ ⟹ d_{N} = sup_{x \in (- 1, 1)} | \frac{x^{N + 1}}{1 - x} | \\ x \to 1 ⟹ \frac{x^{N + 1}}{1 - x} \to \infty ⟹ d_{N} ↛ 0 ⟹ \sum_{n = 0}^{\infty} x^{n} ⇉̸ \frac{1}{1 - x} \end{matrix}

\begin{matrix} x \in [0, \frac{1}{23}] ⟹ d_{N} = sup_{x \in [0, \frac{1}{23}]} | \frac{x^{N + 1}}{1 - x} | = \frac{{(\frac{1}{23})}^{N + 1}}{1 - \frac{1}{23}} \to 0 \end{matrix}

\begin{matrix} Let \sum_{n = 1}^{\infty} f_{n} (x) \\ Let \sum_{n = 1}^{\infty} a_{n} \to M \\ Let \forall n \in N, \forall x \in A : | f_{n} | \leq a_{n} \\ Then \sum_{n = 1}^{\infty} f_{n} (x) ⇉ S (x) \\ And \sum_{n = 1}^{\infty} | f_{n} (x) | ⇉ S (x) \\ Proof: \\ \sum_{n = 1}^{\infty} a_{n} converges ⟹ \forall ε > 0 : \exists N_{ε} : \forall M > N > N_{ε} : S_{M} - S_{N} < \frac{ε}{2} \\ Let ε > 0 \\ \forall x \in A : \forall m > n > N_{ε} : | S_{M} (x) - S_{N} (x) | = | \sum_{n = N + 1}^{M} f_{n} (x) | \leq \sum_{n = N + 1}^{M} | f_{n} (x) | \leq \\ \leq \sum_{n = N + 1}^{M} a_{n} = S_{M} - S_{N} < \frac{ε}{2} \\ Let N > N_{ε} \\ | S (x) - S_{N} (x) | = | lim_{M \to \infty} S_{M} (x) - S_{N} (x) | = lim_{M \to \infty} | S_{M} (x) - S_{N} (x) | \leq \frac{ε}{2} < ε \\ ⟹ S_{N} (x) ⇉ S (x) ⟹ \sum_{n = 1}^{\infty} f_{n} (x) ⇉ S (x) \\ Similar proof for \sum_{n = 1}^{\infty} | f_{n} (x) | ⇉ S (x) \end{matrix}

\begin{matrix} \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \\ ⟹ \sum_{n = 0}^{\infty} n x^{n - 1} = \frac{1}{(1 - x)^{2}} \\ ⟹ \sum_{n = 0}^{\infty} n x^{n} = \frac{x}{(1 - x)^{2}} \\ x = \frac{1}{2} ⟹ \sum_{n = 0}^{\infty} \frac{n}{2^{n}} = \frac{\frac{1}{2}}{{(\frac{1}{2})}^{2}} = 2 \\ Why can we do this? \\ If \\ \sum_{n = 0}^{\infty} (x^{n})^{'} ⇉ g (x), x \in A \\ \exists x_{0} \in A : \sum_{n = 0}^{\infty} x_{0}^{n} \to M \\ Then \sum_{n = 0}^{\infty} (x^{n})^{'} ⇉ {(\frac{1}{1 - x})}^{'} \\ Let A = [0, \frac{1}{2}] \\ \forall x_{0} \in A : \sum_{n = 0}^{\infty} x_{0}^{n} \to M_{x_{0}} \\ \forall n \in N, \forall x \in A : | n x^{n - 1} | = n x^{n - 1} \leq n \cdot {(\frac{1}{2})}^{n - 1} = \frac{n}{2^{n - 1}} \\ lim_{n \to \infty} \sqrt[n]{\frac{n}{2^{n - 1}}} = lim_{n \to \infty} \frac{\sqrt[n]{2} \cdot \sqrt[n]{n}}{\sqrt[n]{2^{n}}} = \frac{1 \cdot 1}{2} = \frac{1}{2} \\ ⟹ \sum_{n = 0}^{\infty} \frac{n}{2^{n - 1}} \to M ⟹ By the Weierstrass M-test \sum_{n = 0}^{\infty} n x^{n - 1} ⇉ g (x) \\ ⟹ \sum_{n = 0}^{\infty} (x^{n})^{'} ⇉ {(\frac{1}{1 - x})}^{'} \end{matrix}