Common tangent problem 3

$\lim_{n \rightarrow \infty} \sum_{j = 1}^{n} (\dfrac{j}{n})^n x^{n - j} = \sum_{j = 0}^{n - 1} (1 - \dfrac{j}{n})^{n} x^{j} = \sum_{n = 0}^{\infty} (\dfrac{x}{e})^n = \dfrac{e}{e - x}$ on $|x| < e$

$\implies \lim_{n \rightarrow \infty} \dfrac{\sum_{j = 1}^{n} x^{j}}{\sum_{j = 1}^{n} (\dfrac{j}{n})^n x^{n - j}} = (\dfrac{x}{1 - x})(\dfrac{e - x}{e}) = \boxed{\dfrac{x(e - x)}{e(1 - x)}}$ on $|x| < 1$

$\implies \dfrac{\sum_{n = 1}^{\infty} (-1)^{n + 1} x^n}{\sum_{n = 0}^{\infty} (-1)^n (\dfrac{x}{e})^n} =$ $\dfrac{-\sum_{n = 1}^{\infty} (-x)^n}{\sum_{n = 0}^{\infty} (\dfrac{-x}{e})^n} =$ $\dfrac{x(e + x)}{e(1 + x)}$ on $|x| < 1$

$\implies (\dfrac{\sum_{j = 1}^{n} (-1)^{n - j} (\dfrac{j}{n})^n x^{n - j})}{\sum_{j = 1}^{n} (-1)^{j + 1} x^{j}}) = \boxed{\dfrac{e(1 + x)}{x(e + x)}}$ on $|x| < 1$

$\implies m(x) = \dfrac{(e - x)(1 + x)}{(1 - x)(e + x)}$

Let $p(x) = ax^2 + bx$ .

$\lim_{x \rightarrow 0} p(x) = \lim_{x \rightarrow 0} m(x) = 0 \implies c = 1$

$m(e - 2) = \dfrac{1}{3 - e} = p(e - 2) = (e - 2)^2 a + b(e -2) + 1 \implies$

$\boxed{(e - 2)a + b = \dfrac{1}{3 - e}}$ .

$\dfrac{dm}{dx} = \dfrac{2(e - 1)}{1 + e}(\dfrac{1}{1 - x)^2} + \dfrac{e}{(e + x)^2})$ and $\dfrac{dp}{dx} = 2ax + b$

$\dfrac{d}{dx}(m(x))|_{x = e - 2} = \dfrac{e^2 - 3e + 4}{2(3 - e)^2 (e - 1)} = \dfrac{d}{dx}(p(x))|_{x = e - 2} = 2(e - 2)a + b \implies$

$\boxed{2(e - 2)a + b = \dfrac{e^2 - 3e + 4}{2(3 - e)^2 (e - 1)}}$

and

$\boxed{(e - 2)a + b = \dfrac{1}{3 - e}}$ .

Solving the system above we obtain:

$a = \dfrac{3e - 5}{2(3 - e)^2(e - 1)}$ and $b = -\dfrac{5e^2 - 19e + 16}{2(3 - e)^2(e - 1)} \implies$

$a + b = -\dfrac{2e^2 - 13e + 13}{2(3 - e)(e - 1)} \approx \boxed{7.80830520}$

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Note: Using point $A: (e - 2, \dfrac{1}{3 - e})$ the equation of the common tangent line is $y - \dfrac{1}{3 - e} = (\dfrac{e^2 - 3e + 4}{2(3 - e)^2(e - 1)})(x + 2 - e)$ .