Yeah, that was a good suggestion indeed, it seems almost trivial now. Let {x} be given by
x_n=1/\sqrt{n} if n=m^2 for some m, 0 else. So we would have x={1,0,0,1/2,0,0,0,0,1/3,0, ...} for the first few terms. Then:
\sum_{n=1}^\infty x_n^2 = \sum_{m=1}^\infty (1/\sqrt{m^2})^2=\sum...