Special Series — Question 7

*Question 7 Evaluate $\displaystyle\sum_{n=1}^{\infty}\frac1{(2n)^2}$.

1. Factor the constant scale from the summand.

2. Use the Basel sum to evaluate the series.

3. Relate the answer to the odd-square contribution and the full series.

*Question 7 – Solution

Since $(2n)^2=4n^2$, $$\frac1{(2n)^2}=\frac14\frac1{n^2}.$$

The Basel series converges, so its constant factor may be pulled outside: $$\sum_{n=1}^{\infty}\frac1{(2n)^2} =\frac14\sum_{n=1}^{\infty}\frac1{n^2} =\frac14\cdot\frac{\pi^2}{6} =\boxed{\frac{\pi^2}{24}}.$$

The substitution $m=2n$ selects precisely the even-indexed Basel terms. Their total is one-fourth of the complete sum because squaring the index contributes the scale factor $2^2=4$.

Adding the odd-square value $\pi^2/8$ gives $$\frac{\pi^2}{24}+\frac{\pi^2}{8} =\frac{\pi^2}{6},$$ recovering the full Basel sum.