Special Series — Question 6

*Question 6 Evaluate $\displaystyle\sum_{n=1}^{\infty}\frac1{(2n-1)^2}$.

1. Explain why this selects exactly the odd-indexed terms of the Basel series.

2. Compute the even-indexed contribution by scaling $\zeta(2)$.

3. Subtract to obtain the odd-square sum and verify that the two pieces recombine correctly.

*Question 6 – Solution

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

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

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

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.