Just seems infinite

Start with $1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6}.$

Then if we divide both sides by $2^2 = 4$ , then we find $\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \frac{1}{8^2} + \cdots = \frac{\pi^2}{24}$

Then, subtracting $2\times$ this infinite series from the first one, we have $1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \cdots = \frac{\pi^2}{6} - 2\left(\frac{\pi^2}{24}\right) = \frac{\pi^2}{12}.$