Common Tangent Mania

Let $|x| > 1$ .

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

$\implies f(x) = \lim_{n \rightarrow \infty} \dfrac{\sum_{j = 1}^{n} (\dfrac{1}{x})^{j}}{\sum_{j = 1}^{n} (\dfrac{j}{n})^n (\dfrac{1}{x})^{n - j}} = \dfrac{ex - 1}{ex(x - 1)}$ .on $|x| > 1$ .

$f(e) = g(e) \implies e^2a + eb = \dfrac{e + 1}{e^2} \implies \boxed{ea + b = \dfrac{e + 1}{e^3}}$

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

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

$\boxed{ea + b = \dfrac{e + 1}{e^3}}$

Solving the system we obtain:

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

$\implies a + b = \dfrac{3e^2 + 2e - 2}{e^4} \approx \boxed{0.4689487}$ .

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Note: The common tangent line is: $(e^2 + e - 1)x + e^3(e - 1)y - e(2e^2 + e - 2) = 0$ .