I Was Very Amazed At The Solution 11

$\begin{aligned} S(x) & = \sum_{n=1}^\infty \frac {(-1)^{n+1}x^{2n-1}}{(2n-1)(n-1)!} \\ & = \frac {x}{1 \cdot 0!} - \frac {x^3}{3 \cdot 1!} + \frac {x^5}{5 \cdot 2!} - \cdots \\ & = \int e^{-x^2} dx \\ & = \frac {\sqrt \pi}2 \text{ erf }(x) + C & \small \color{#3D99F6} \text{where erf }(x) \text{ is the error function and } C \text{, the constant of integration.} \\ & = \frac {\sqrt \pi}2 \text{ erf }(x) & \small \color{#3D99F6} \text{Since }S(0) = 0, \text{ erf }(0) = 0, \implies C = 0 \end{aligned}$

Therefore,

$\begin{aligned} X & = \left[ \frac 1\pi \left(\lim_{x \to \infty} \sum_{n=1}^\infty \frac {(-1)^{n+1}x^{2n-1}}{(2n-1)(n-1)!} \right)^2\right]^{-1} \\ & = \left[ \frac 1\pi \left(\lim_{x \to \infty} \frac {\sqrt \pi}2 \text{ erf }(x) \right)^2\right]^{-1} & \small \color{#3D99F6} \text{Note that erf }(\infty) = 1 \\ & = \left[ \frac 1\pi \left(\frac {\sqrt \pi}2 \right)^2\right]^{-1} \\ & = \boxed{4} \end{aligned}$

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Reference: Error function .