Tangents and Series

Let $|x| > 1$ .

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

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

and

$g(x) = \dfrac{\sum_{n = 1}^{\infty} (-1)^{ n + 1} (\dfrac{1}{x})^n}{\sum_{n = 0}^{\infty} (-1)^n (\dfrac{1}{ex})^n} = \dfrac{-\sum_{n = 1}^{\infty} (\dfrac{1}{-x})^n}{\sum_{n = 0}^{\infty} (\dfrac{1}{e(-x)})^n} =$ $\boxed{\dfrac{ex + 1}{ex(x + 1)}}$ on $|x| > 1$ .

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

$a > 0 \implies \dfrac{d}{dx}(f(x))|_{x = a} = -\dfrac{ea^2 - 2a + 1}{ea^2(a - 1)^2} = \dfrac{d}{dx}(g(x))|_{x = -a}$

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

The tangent line to the curve $f(x)$ at $A: (e, \dfrac{e^2 - 1}{e^2(e - 1)})$ is:

$(e^3 - 2e + 1)x + e^3(e - 1)^2y - (2e^4 - e^3 - 3e^2 + 2e) = 0$

and

The tangent line to the curve $g(x)$ at $B: (-e, -\dfrac{e^2 - 1}{e^2(e - 1)})$ is:

$(e^3 - 2e + 1)x + e^3(e - 1)^2y + (2e^4 - e^3 - 3e^2 + 2e) = 0$ .

Using point $B: (-e, -\dfrac{e^2 - 1}{e^2(e - 1)})$ and the line $(e^3 - 2e + 1)x + e^3(e - 1)^2y - (2e^4 - e^3 - 3e^2 + 2e) = 0$ the distance $d = BC = \dfrac{|2e(-2e^3 + e^2 + 3e - 2)|}{(e - 1)\sqrt{e^8 - 2e^7 + e^6 + e^4 + 2e^3 - e^2 - 2e + 1}} \approx \boxed{2.36026}$