Convergence of a Related Series (3.7)

Walter Rudin Principles of Mathematical Analysis, chapter 3, exercise 7:

Let $a_n≥0$. Show that the convergence of $∑a_n$ implies the convergence of $$\sum \dfrac{\sqrt{a_n}}{n}$$
Proof:

Lemma: If $a_n≥a_{n+1}≥0$ and $∑a_n$ converges, then $∑\sqrt{a_{2^n}}$ converges. Proof: By theorem 3.25, $∑2^na_{2^n}$ converges, so by the root test $\text{lim sup }\sqrt[n]{2^na_{2^n}}=2\text{lim sup }\sqrt[n]{a_{2^n}}≤1$ so $\text{lim sup }\sqrt[n]{a_{2^n}}≤1/2$ implying $∑\sqrt{a_{2^n}}$ converges by the root test by $\text{lim sup }\sqrt[n]{\sqrt{a_{2^n}}}≤\sqrt{1/2} < 1$ (since one may show without advancing in theory that $\text{lim sup }\sqrt{x_n}≤\sqrt{\text{lim sup }x_n}$ for any positive sequence $x_n$).

Now, every nonempty subset of a convergent series' terms $\{a_n\}$ has an absolute greatest element, since by the epsilon-delta method there are only finitely many elements absolutely greater than any chosen term. Since $∑a_n=∑|a_n|$, use this to inductively construct a rearrangement $∑b_n$ such that $b_n≥b_{n+1}$ without losing convergence.

If we prove $∑\dfrac{\sqrt{b_n}}{n}$ converges, then though $∑\dfrac{\sqrt{a_n}}{n}$ is not generally a rearrangement, we may observe the effect of transpositions on partial sums to show the former bounds the latter, demonstrating the latter's convergence. So by 3.25, the former is equivalent to $∑\sqrt{b_{2^n}}$ converging, which is established by the lemma.$~\square$