Special Series — Question 6

Question 6 – Solution

Step 1: Partition the Basel series.

Every positive integer is either even or odd, and the terms are positive, so $$\sum_{m=1}^{\infty}\frac1{m^2} =\sum_{n=1}^{\infty}\frac1{(2n)^2} +\sum_{n=1}^{\infty}\frac1{(2n-1)^2}.$$

Step 2: Evaluate the even contribution.

Using Euler’s value $\zeta(2)=\pi^2/6$, $$\sum_{n=1}^{\infty}\frac1{(2n)^2} =\frac14\sum_{n=1}^{\infty}\frac1{n^2} =\frac{\pi^2}{24}.$$

Step 3: Subtract to isolate the odd terms.

$$\sum_{n=1}^{\infty}\frac1{(2n-1)^2} =\frac{\pi^2}{6}-\frac{\pi^2}{24} =\boxed{\frac{\pi^2}{8}}.$$

Step 4: Verify the decomposition.

The even portion is one-fourth of the full sum, while the odd portion is three-fourths. Their sum is $(1/4+3/4)\zeta(2)=\zeta(2)$, as required.