More on Alternating Series

Note: You can check that both series above converge on $[-1,1]$ .

Let $h(x) = \sum_{n = 2}^{\infty} \dfrac{x^n}{n} \implies \dfrac{d}{dx}(h(x)) = \sum_{n = 2}^{\infty} x^{n - 1} = \dfrac{x}{1 - x} \implies h(x) = \int_{0}^{x} \dfrac{x}{1 - x} dx =$ $\int_{0}^{x} (\dfrac{1}{1 - x} - 1) dx =\ln(\dfrac{1}{1 - x}) - x$ on $|x| < 1$ .

Let $(|x| \leq 1)$ .

Let $f(x) = \sum_{n = 2}^{\infty} \dfrac{x^{n - 1}}{(n - 1)n} \implies \dfrac{d}{dx}(f(x)) =\sum_{n = 2}^{\infty} \dfrac{x^{n - 2}}{n} = \dfrac{1}{x^2}\sum_{n = 2}^{\infty} \dfrac{x^{n}}{n} =$ $\dfrac{1}{x^2}(\ln(\dfrac{1}{1 - x}) - x) \implies f(x) = \int \dfrac{1}{x^2}(\ln(\dfrac{1}{1 - x}) - x) dx$

Let $u = \ln(\dfrac{1}{1 - x}) - x, du = \dfrac{x}{1 - x}$ and $dv = x^{-2} dx \implies v = \dfrac{-1}{x}$

$\implies \boxed{f(x) = \dfrac{-1}{x}(\ln(\dfrac{1}{1 - x}) - x))|_{\epsilon}^{x} + \int_{\epsilon}^{x} \dfrac{1}{1 - x} dx = \dfrac{(1 - x)\ln(1 - x) + x}{x}}$ , where $\lim_{\epsilon \rightarrow 0} \dfrac{(1 - \epsilon)\ln(1 - \epsilon) + \epsilon}{\epsilon} = 0$ .

$\boxed{g(x) = \sum_{n = 2}^{\infty} (-1)^{n} \dfrac{x^{n - 1}}{(n - 1)n} = - \sum_{n = 2}^{\infty} \dfrac{(-x)^{n - 1}}{(n - 1)n} = -(\dfrac{(1 + x)\ln(1 + x) - x}{-x}) = \dfrac{(1 + x)\ln(1 + x) - x}{x}}$ .

$\dfrac{d}{dx}(g(x)) = \dfrac{1}{x} - \dfrac{\ln(1 + x)}{x^2}$ and $\dfrac{d}{dx}(f(x)) = \dfrac{-1}{x} - \dfrac{\ln(1 - x)}{x^2}$

Let $(0 < a \leq 1)$ .

$\dfrac{d}{dx}(g(x))|_{x = a} = \dfrac{1}{a} - \dfrac{\ln(1 + a)}{a^2} = \dfrac{d}{dx}(f(x))|_{x = -a}$

$\implies \overleftrightarrow{AC}$ is tangent to $g(x)$ at $A: (a, \dfrac{(1 + a)\ln(1 + a) - a}{a})$ and $\overleftrightarrow{BD}$ is tangent to $f(x)$ at $B:( -a, \dfrac{a - (1 + a)\ln(1 + a)}{a})$

$\implies y = (\dfrac{1}{a} - \dfrac{\ln(1 + a)}{a^2})x + \dfrac{(2 + a)\ln(1 + a) - 2a}{a}$ and $y = (\dfrac{1}{a} - \dfrac{\ln(1 + a)}{a^2})x - \dfrac{(2 + a)\ln(1 + a) - 2a}{a}$ respectively.

Letting $a = 1 \implies y = (1 - \ln(2))x + (3\ln(2) - 2)$ and $y = (1 - \ln(2))x - (3\ln(2) - 2)$

Using $B: (-1,1 - 2\ln(2)) \implies m_{\perp} = \dfrac{-1}{1 - \ln(2)} \implies$ the normal line to $f(x)$ at $x = -1$ is $y = (\dfrac{-1}{1 - \ln(2)})x + \dfrac{2(\ln(2))^2 - 3\ln(2)}{1 - \ln(2)}$

$\implies$

$y + (\dfrac{1}{1 - \ln(2)})x = \dfrac{2(\ln(2))^2 - 3\ln(2)}{1 - \ln(2)}$

$y - (1 - \ln(2))x = (3\ln(2) - 2)$

$\implies x_{0} = \dfrac{5(\ln(2))^2 - 8\ln(2) + 2}{(\ln(2) - 1)^2 + 1}$ and $y_{0} = \dfrac{-2(\ln(2))^3 + 5(\ln(2))^2 - 2}{(\ln(2) - 1)^2 + 1}$

Using $C:(x_{0},y_{0})$ and $B: (-1,1 - 2\ln(2))$ $\implies \triangle{x} = \dfrac{2(3\ln(2) - 2)(\ln(2) - 1)}{(\ln(2) - 1)^2 + 1}$ and $\triangle{y} = \dfrac{2(3\ln(2) - 2)}{(\ln(2) - 1)^2 + 1}$

$\implies |BC| = \dfrac{2(3\ln2 -2)}{\sqrt{(\ln2 -1)^2 +1}} \approx \boxed{0.15189291}$