 Solution

Most of the mean-square signal power lies at low frequencies. Each nonzero coefficient has magnitude $2/(\pi |r|)\mathrm{V}$, so its power contribution decreases as $r^{-2}$. The pair at the fundamental frequencies $\pm 0.25\mathrm{Hz}$ alone containsof the total. The first three positive odd harmonics, together with their negative-frequency partners, contain about $93.3\%$. The harmonic power of a square wave still has a nonzero tail: higher harmonics are needed to reproduce sharp edges even though their combined power is small.