$l^2$ convergence

This follows from Hardy's inequality. Looking at its proof, you see that Cauchy-Schwarz IS the right way to go, applied to $$ \sum_{j=1}^n y_j = \sum_{j=1}^n j^{-1/4} j^{1/4}y_j. $$

To give intuition why you first write the sum like this: Cauchy-Schwarz is sharp when two functions are multiples of each other. If you apply it simply to $fg$, with $f(j)=1, g(j)=y_j$ then they are less and less multiples of each others as $n\to \infty$ because $y_j$ converges to zero and $1$ does not. With the choice above, $f(j)=j^{-1/4}$, $g(j)=j^{1/4}y_j$, both are appropriately multiples, at least at infinity. This is of course very handwaving, but if you consider $y_j=j^{-1/2}$, which is at the very edge of being (not) in $\ell^2$, then $f$ and $g$ above are actually the same