Infinite cuts and joints in Infinite disks

Let the mass per unit area be $m$ . The mass and location of X coordinate of the COM of the largest circle is:

$M_0 = m (\pi R^2) \ ; x_0 = 0$ $R =10$

Similarly, the mass and location of the X coordinate of the COM of the second-largest circle is:

$M_1 = m\left(\frac{\pi R^2}{2^2}\right) \ ; \ x_1 = R - \frac{R}{2}$

That of the third-largest circle:

$M_2 = m\left(\frac{\pi R^2}{4^2}\right) \ ; \ x_2 = R - \frac{R}{4}$

And so on. Generally, for the n-th largest circle:

$M_{n} = m\left(\frac{\pi R^2}{2^{2n}}\right) \ ; \ x_{n-1} = R - \frac{R}{2^{n}}$

The odd subscripts correspond to the yellow disks that are cut out and the even subscripts correspond to the red disks added. The X coordinate of the COM of this entire system is, therefore:

$X_{COM} = \frac{M_0x_0 - M_1x_1 + M_2x_2 - M_3x_3 + M_4x_4 - M_5x_5 + M_6x_6 - \dots}{M_0-M_1 +M_2-M_3+M_4 \dots}=\frac{N}{D}$

Where $N$ and $D$ can be generally written as: $N = \sum_{k=0}^{\infty} (-1)^{k} M_{k} x_{k}$ $D = \sum_{k=0}^{\infty} (-1)^{k} M_{k}$

$\implies N = \sum_{k=0}^{\infty} (-1)^k m\left(\frac{\pi R^2}{2^{2k}}\right) \left( R - \frac{R}{2^{k}} \right)$ $\implies D = \sum_{k=0}^{\infty} (-1)^k m\left(\frac{\pi R^2}{2^{2k}}\right)$

I am leaving out the steps of the final calculations. The answer is $X_{COM} \approx -1.111$ . Since the location of the COM of the largest full circle is the origin, the required answer is $\approx \boxed{1.111}$