Prob. 7, Chap. 3 in Baby Rudin: If $a_n \geq 0$, then how does convergence of $\sum a_n$ imply convergence of $\sum \frac{\sqrt{a_n}}{n}$?

Question

Here's Prob. 7 in the Exercises of Chapter 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Prove that the convergence of $\sum a_n$ implies the convergence of $$ \sum \frac{\sqrt{a_n}}{n}$$ if $a_n \geq 0$.

How to show this? I have no clue! Can anybody here please be of any help?

Is this result a special case (or application) of a more general result about infinite series?

Answer 1

By the Cauchy-Schwarz inequality, $$ \sum\frac{\sqrt{a_n}}n\le\biggl(\sum a_n\biggr)^{1/2}\biggl(\sum\frac1{n^2}\biggr)^{1/2}. $$

Answer 2

We need to establish that the series $\sum\limits_{n=1}^{\infty} \frac{\sqrt{a_n}}{n}$ is convergent where it is given that the series of non-negative terms $\sum\limits_{n=1}^{\infty} a_n$ is convergent.

here we see that $\sum a_n$ is convergent implies $\sum a_n^2$ is also convergent. Now $\sum a_n^2 +\sum \frac{1}{n^2}=\sum (a_n^2+\frac{1}{n^2})\geq \sum 2\sqrt{a_n^2\cdot \frac{1}{n^2}}=2\sum \frac{a_n}{n}$.

The LHS of this inequality is bounded above. hence the proof.