Is this correct for Rudin exercise 3.7? Prove the series is convergent
Solution 1:
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 $\sum\limits_{k=n}^m a_k$.