Splitting the sum of 1/n² into even and odd terms

Why splitting is legitimate. For an absolutely convergent series, any grouping or rearrangement of the terms leaves the sum unchanged.

Even terms. $\sum_{k\ge1}\frac1{(2k)^2}=\frac14\sum_{k\ge1}\frac1{k^2}=\frac{\pi^2}{24}$.

Odd terms. $A=\frac{\pi^2}6-\frac{\pi^2}{24}=\dfrac{\pi^2}8.$

Alternating sum. $B=A-\frac{\pi^2}{24}=\frac{\pi^2}8-\frac{\pi^2}{24}=\dfrac{\pi^2}{12}.$