Infi-2 17

Uniform convergence of power series #lemma

\begin{matrix} \sum_{n = 0}^{\infty} a_{n} (x - a)^{n} converges at least pointwise on (a - R, a + R) \\ By Weierstrass M-test, absolute convergence implies uniform convergence \\ ⟹ \forall [b, c] \subset (a - R, a + R) : \sum_{n = 0}^{\infty} a_{n} (x - a)^{n} converges uniformly on [b, c] \end{matrix}

\begin{matrix} Differentiation and integration of power series does not affect the convergence radius \end{matrix}

\begin{matrix} \sum_{n = 0}^{\infty} \frac{n}{2^{n}} = \sum_{n = 0}^{\infty} n {(\frac{1}{2})}^{n} \\ \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}} \\ ⟹ \sum_{n = 0}^{\infty} n {(\frac{1}{2})}^{n} = \frac{\frac{1}{2}}{{(1 - (\frac{1}{2}))}^{2}} = 2 \end{matrix}

\begin{matrix} \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} x \in (- 1, 1) \\ ⟹ \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1} = - \ln | 1 - x | \\ Series converges for x = - 1 \\ But does it converge to - \ln | 1 - x | ? \\ Turns out yes! \\ S (- 1) = lim_{x \to - 1} S (x) = lim_{x \to - 1} - \ln | 1 - x | = - \ln (2) \end{matrix}

\begin{matrix} Power series converges uniformly on edges of convergence interval too! \end{matrix}

\begin{matrix} Equality "series = function" allows us to calculate sums \end{matrix}

\begin{matrix} e^{x} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} x \in R \\ e = \sum_{n = 0}^{\infty} \frac{1}{n!} \end{matrix}