Metric space question?

If a_n is a sequence greater than 0, and the sum of a_n converges, then the sum (n=1 to infinity) of sqrt(a_n)/n converges.

I had a similar problem to this one a few days ago. If we can show that both sqrt(a_n) and 1/n are both in L^2 (L^2 is the typical definition, defined as {s_n}(n=1 to infinity) : sum (n=1 to infinity) s_n^2 < infinity)

Since we know that a_n converges, and that it is obviously sqrt(a_n)^2, so sqrt(a_n) is in L^2., and (1/n)^2 = 1/n^2, and we know that that sums to pi^2/6, so 1/n is also in L^2. So, by Schwarz inequality, the product of them is absolutely convergent, and thus convergent as well.