Show that 4/9 times series of basel problem can be written as following sum

Solution 1:

Note that:

$$\frac{1}{m^2}s=\sum_{k=1}^\infty\frac{1}{(mk)^2}$$

Then we can eliminate thirds as follows, and call it a different series by name $s'$:

$$s'=s-\frac{1}{9}s=\sum_{k=1,k\neq3n,n\in\Bbb N}\frac{1}{k^2}$$

The evens are:

$$\frac{1}{4}s'=\sum_{k\text{ even, not mult. $3$}}\frac{1}{k^2}$$

To keep the odds, and make the evens negative, I must first subtract all the evens, leaving a sum of only odds, and then subtract all the evens again to obtain positive odds and negative evens:

$$s'-\frac{1}{4}s'-\frac{1}{4}s'=\frac{1}{2}s'=1-\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{5^2}+\cdots$$

But $s'=s-\frac{1}{9}s=\frac{8}{9}s$, so your desired sum is $\frac{1}{2}s'=\frac{1}{2}\cdot\frac{8}{9}s=\frac{4}{9}s$.