Alternating series 3

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

Let $h(x) = \sum_{n = 3}^{\infty} \dfrac{x^n}{n} \implies \dfrac{d}{dx}(h(x)) = \sum_{n = 3}^{\infty} x^{n - 1} = \dfrac{x^2}{1 - x}$ .

Let $u = 1 -x \implies du - -dx \implies h(x) = \int_{0}^{x} \dfrac{x^2}{1 - x} dx =$ $\int_{1}^{1 - x} (\dfrac{-1}{u} + 2 - u) du = \ln(\dfrac{1}{u}) + 2u - \dfrac{u^2}{2}|_{1}^{1 - x} = \ln(\dfrac{1}{1 - x}) - x - \dfrac{x^2}{2}$ on $|x| < 1$ .

Let $w(x) = \sum_{n = 3}^{\infty} \dfrac{x^{n - 1}}{(n - 1)n} \implies \dfrac{d}{dx}(w(x)) =$ $\sum_{n = 3}^{\infty} \dfrac{x^{n - 2}}{n} = \dfrac{1}{x^2}\sum_{n = 3}^{\infty} \dfrac{x^n}{n} =$ $\dfrac{1}{x^2}(\ln(\dfrac{1}{1 - x}) - x) - \dfrac{1}{2} \implies$ $w(x) = \int_{\epsilon}^{x} \dfrac{1}{x^2}(\ln(\dfrac{1}{1 - x}) - x) - \dfrac{1}{2} \:\ dx$ .

Let $I(x) = \int_{\epsilon}^{x} \dfrac{1}{x^2}(\ln(\dfrac{1}{1 - x}) - x) dx$

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

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

$\therefore \boxed{w(x) = \int_{\epsilon}^{x} \dfrac{1}{x^2}(\ln(\dfrac{1}{1 - x}) - x) - \dfrac{1}{2} \:\ dx = \dfrac{(1 - x)\ln(1 - x) + x}{x} - \dfrac{x}{2}}$ on $|x| \leq 1$ .

Let $g(x) = \sum_{n = 3}^{\infty} \dfrac{x^{n - 2}}{(n - 2)(n - 1)n} \implies \dfrac{d}{dx}(g(x)) = \sum_{n = 3}^{\infty} \dfrac{x^{n - 3}}{(n - 1)n} = \dfrac{1}{x^2}\sum_{n = 3}^{\infty} \dfrac{x^{n - 1}}{(n - 1)n}$ = $\dfrac{(1 - x)\ln(1 - x) + x}{x^3} - \dfrac{1}{2x}$ .

Let $I^{*}(x) = \int \dfrac{(1 - x)\ln(1 - x) + x}{x^3} \:\ dx$ .

Let $u = (1 - x)\ln(1 - x) + x, du = -\ln(1 - x), dv = x^{-3}$ and $v = \dfrac{-1}{2x^2} \implies$

$I(x) = \dfrac{-1}{2x^2}((1 - x)\ln(1 - x) + x) - \dfrac{1}{2}\int \dfrac{\ln(1 - x)}{x^2}$ .

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

$\int \dfrac{\ln(1 - x)}{x^2} dx = \dfrac{-1}{x}\ln(1 - x) - \int \dfrac{1}{x} + \dfrac{1}{1 - x} dx \implies$

$I^{*}(x) = \dfrac{1}{2}(\dfrac{1 - x}{x}\ln(1 - x) + \ln(x)) \implies$ $g(x) = \dfrac{-1}{2x^2}((1 - x)\ln(1 - x) + x) + \dfrac{1}{2}((\dfrac{1 - x}{x})\ln(1 - x) + \ln(x)) - \dfrac{1}{2}\ln(x)|_{\epsilon}^{x} =$ $-(\dfrac{(x - 1)^2\ln(1 - x) + x}{2x^2})|_{\epsilon}^{x} = -\dfrac{(x - 1)^2\ln(1 - x) + x}{2x^2} - \dfrac{3}{4} =$ $\dfrac{-2(x - 1)^2\ln(1 - x) - 2x + 3x^2}{4x^2}$ , where $\dfrac{-1}{2}\lim_{\epsilon \rightarrow 0} \dfrac{(\epsilon - 1)^2\ln(1 - \epsilon) + \epsilon}{\epsilon^2} = \dfrac{-3}{4}$ and $|x| \leq 1$

$\implies f(x) = \sum_{n = 3}^{\infty} (-1)^{n + 1} \dfrac{x^{n - 2}}{(n - 2)(n - 1)n} =$ $-\sum_{n = 3}^{\infty} \dfrac{(-x)^{n - 2}}{(n - 2)(n - 1)n} = \dfrac{2(x + 1)^2\ln(x + 1) - 2x - 3x^2}{4x^2}$ on $|x| \leq 1$ .

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

and,

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

$\implies \dfrac{d}{dx}(f(x))|_{x = a} = \dfrac{1}{2}(\dfrac{a^2 + 2a - 2(a + 1)\ln(1 + a)}{a^3}) =$ $\dfrac{d}{dx}(g(x))|_{x = -a}$

$\implies \overleftrightarrow{AC}$ is tangent to $f(x)$ at $A: (\dfrac{2(a + 1)^2\ln(a + 1) - 2a - 3a^2}{4a^2})$ and $\overleftrightarrow{BD}$ is tangent to $g(x)$ at $B:( -a, \dfrac{-2(a + 1)^2\ln(a + 1) + 2a + 3a^2}{4a^2} )$

Letting $a = 1 \implies A: (1, \dfrac{8\ln(2) - 5}{4})$ and $B: (-1, \dfrac{5 - 8\ln(2)}{4}) \implies$ $y = \dfrac{1}{2}(3 -4\ln(2))x + \dfrac{16\ln(2) - 11}{4}$ and $y = \dfrac{1}{2}(3 -4\ln(2))x - \dfrac{16\ln(2) - 11}{4}$ .

Using $B: (-1, \dfrac{5 - 8\ln(2)}{4}) \implies m_{\perp} = \dfrac{-2}{3 - 4\ln(2)} \implies$ the normal line to $g(x)$ at $x = -1$ is $y = \dfrac{-2}{3 - 4\ln(2)}x + \dfrac{32(\ln(2))^2 - 44\ln(2) + 7}{4(3 - 4\ln(2)}$ .

To find point $C: (x_{0},y_{0})$ we solve the system:

$4y - 2(3 - 4\ln(2))x = 16\ln(2) - 11$

$4(3 - 4\ln(2))y + 8x = 32(\ln(2))^2 - 44\ln(2) + 7$

Solving the system above we obtain:

$x_{0} = \dfrac{4(12(\ln(2))^2 - 17\ln(2) + 5)}{16(\ln(2))^2 -24\ln(2) + 13}$

and

$y_{0} = \dfrac{-128(\ln(2))^3 + 272(\ln(2))^2 - 96\ln(2) - 23}{4(16(\ln(2))^2 -24\ln(2) + 13)}$

Using $C: (x_{0}, y_{0})$ and $B: (-1, \dfrac{5 - 8\ln(2)}{4}) \implies \triangle{x} = \dfrac{64(\ln(2))^2 - 92\ln(2) + 33}{16(\ln(2))^2 - 24\ln(2) + 13}$ and $\triangle{y} = \dfrac{32\ln(2) - 22}{16(\ln(2))^2 - 24\ln(2) + 13}$

$\implies BC = \dfrac{16\ln(2) - 11}{\sqrt{16(\ln(2))^2 - 24\ln(2) + 13}} \approx \boxed{0.044888}$ .

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Bonus:

Let $(0 < a \leq 1)$ and $j$ be a odd positive integer and $|x| \leq 1$ .

Let $w(x) = \sum_{n = j + 2}^{\infty} \dfrac{x^{n - j}}{(n - j)(n - j + 1) \cdot \cdot \cdot (n)}$

$g(x) = \sum_{n = j + 2}^{\infty} \dfrac{x^{n - j - 1}}{(n - j - 1)(n - j) \cdot \cdot \cdot (n)} \implies$
$\dfrac{d}{dx}(g(x)) = \sum_{n = j + 2}^{\infty} \dfrac{x^{n - j - 2}}{(n - j) \cdot \cdot \cdot (n)} =$ $\dfrac{1}{x^2} \sum_{n = j + 1}^{\infty} \dfrac{x^{n - j}}{(n - j)(n - j + 1) \cdot \cdot \cdot (n)} = \dfrac{1}{x^2} w(x)$

$\implies g(x) = \lim_{\epsilon \rightarrow 0}\int_{\epsilon}^{x} \dfrac{1}{x^2} w(x)$ .

$f(x) = \sum_{n = j + 2}^{\infty} (-1)^{n + 1} \dfrac{x^{n - j - 1}}{(n - j - 1)(n - j) \cdot \cdot \cdot (n)} =$ $-\sum_{n = j + 2}^{\infty} \dfrac{(-x)^{n - j - 1}}{(n - j - 1)(n - j) \cdot \cdot \cdot (n)} =$ $-g(-x) \implies \boxed{f(a) = -g(-a)}$

$\dfrac{d}{dx}(f(x))|_{x = a} = \sum_{n = j + 2}^{\infty} (-1)^{n + 1} \dfrac{a^{n - j - 2}}{(n - j)(n - j + 1) \cdot \cdot \cdot (n)} =$ $\dfrac{1}{a^2} \sum_{n = j + 2}^{\infty} (-1)^{n + 1} \dfrac{a^{n - j}}{(n - j)(n - j + 1) \cdot \cdot \cdot (n)} =$ $\dfrac{1}{a^2} \sum_{n = j + 2}^{\infty} \dfrac{(-a)^{n - j}}{(n - j)(n - j + 1) \cdot \cdot \cdot (n)} =$ $\dfrac{1}{a^2} w(-a) = \dfrac{d}{dx}(g(x))|_{x = -a}$

$\therefore \boxed{\dfrac{d}{dx}(f(x))|_{x = a} = \dfrac{d}{dx}(g(x))|_{x = -a}}$ .