So, in baby Rudin we have the exercise
Prove that the convergence of $\sum a_n$ implies the convergence of $$\sum \frac{\sqrt{a_n}}{n}$$ if $a_n \geq 0$
My Approach
Assume $\Sigma a_n$ converges and allow $S_n = \sum_{i=1}^{n} \frac{\sqrt{a_i}}{i}$.
By Cauchy-Schwarz we have $S_n^2 = (\sum_{i=1}^{n} \frac{\sqrt{a_i}}{i})^2 \leq \sum_{i=1}^{n}\frac{1}{i^2} \sum_{i=1}^{n} a_i $. Thus $\lim_{n\to\infty} S_n^2$ is finite. Consequently, $S_n$ converges.
Is this correct? I'm not sure if that's proper use of Cauchy-Schwarz or if I can apply limits like that. I appreciate the feedback.
Your proof is indeed correct.
For the sake of variety, I would suggest an alternative to Cauchy—Schwarz, namely the Am-GM inequality: analogously to what you have written, for any $n\geq 0$, $$ 0\leq \sum_{k=1}^n \frac{\sqrt{a_k}}{k} \stackrel{\tiny\rm (AM-GM)}{\leq} \sum_{k=1}^n \frac{a_k+\frac{1}{k^2}}{2} = \frac{1}{2}\left( \sum_{k=1}^n a_k + \sum_{k=1}^n \frac{1}{k^2}\right) \leq \frac{1}{2}\left( \sum_{k=1}^\infty a_k + \sum_{k=1}^\infty \frac{1}{k^2}\right) $$ Thus, the partial sums $S_n$ are bounding, and non-decreasing: $(S_n)_n$ converges by the monotone convergence theorem.