I think that I have found a way to close your gap. I will let you take a look at it before I send you my bank information for my half of the prize money. My idea is to use the geometric sum and combine it with a particular value of the Riemann zeta function.

Indeed we have

$$\frac{1}{1+x} = 1-x+x^2-x^3+x^4- \dots $$

In particular (for $x = 1$),

$$\frac{1}{2} = 1-1+1-1+1 - \dots = (1-1)\left[1+1+1+1+1+\dots\right] = (1-1) \zeta(0) \, .$$

Since

$$\zeta(s) = 1 + \frac{1}{2^s}+ \frac{1}{3^s}+ \frac{1}{4^s}+ \frac{1}{5^s}+ \dots$$

Then, using the fact that $\zeta(0) = -1/2$ we can write

$$\frac{X-X}{X} = (1-1) = \frac{1}{2\zeta(0)} = -1\, .$$

This is clearly incompatible with $X-X = 0$.