This is Baby Rudin exercise 7 of Chapter 3. Prove that the convergence of $\sum{a_n}$ implies the convergence of $\sum{\sqrt {a_k} \over k}$ if $a_n \ge 0$.
Proof:
I will attempt to show that the cauchy criterion is met so $${\sum_{k=n}^m {\sqrt {a_k} \over k}} \le \epsilon $$ for all $n,m \le N_1$. Im disregarding the absolute value since its all positive terms.
Since $\sum{a_n}$ converges, we know that the $\lim \limits_{n \to \infty} a_n =0$ So there exists some $N_2$ such that for all $n \ge N_2$, $$ a_n \lt \ {{(m-n)}^2k^2(\epsilon)^2} $$ I am again disregarding the absolute values because its always positive Let N=max{$N_1,N_2$} then for all $n,m \ge N$ $$\sum_{k=n}^m {\sqrt {a_k} \over k} \le \sum_{k=n}^m {\sqrt {{(m-n)}^2k^2(\epsilon)^2} \over k} \le \sum_{k=n}^m{(m-n)(\epsilon)} = \epsilon $$
Maybe this is just crazy, should I just use Cauchy-Schwarz inequality and forget about this or could I possibly make this method work?
Best Answer
How about trying Cauchy-Schwarz: $$ \sum_{n=1}^\infty\frac{\sqrt{a_n}}n \le\left(\sum_{n=1}^\infty a_n\right)^{1/2} \left(\sum_{n=1}^\infty\frac1{n^2}\right)^{1/2} $$
You can use Cauchy-Schwarz to make the estimate in your question. $$ \begin{align} \sum_{k=n}^m\frac{\sqrt{a_k}}k &\le\left(\sum_{k=n}^m a_k\right)^{1/2} \left(\sum_{k=n}^m\frac1{k^2}\right)^{1/2}\\ &\le\frac\pi{\sqrt6}\left(\sum_{k=n}^m a_k\right)^{1/2} \end{align} $$ which can be made small by the convergence of $\sum\limits_{k=n}^m a_k$.