# Convergent series exercise from Little Rudin

Problem:

Suppose that for every $n\in\mathbb{N}$, $a_n\in\mathbb{R}$ and $a_n\ge 0$. Given that $$\sum_0^\infty a_n$$ converges, show that $$\sum_1^\infty \frac{\sqrt{a_n}}{n}$$ converges.

Source: Rudin, Principles of Mathematical Analysis, Chapter 3, Exercise 7.

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See here for a proof. –  Fermé somme Jul 1 '14 at 8:40

The Cauchy-Schwarz inequality gives $$\sum_{n=1}^\infty \frac{\sqrt{a_n}}{n}\leq \sqrt{\sum_{n=1}^\infty a_n}\,\sqrt{\sum_{n=1}^\infty \frac{1}{n^2}}<\infty.$$
We have for all real numbers $2ab\leq a^2+b^2$ hence $$0\leq \frac{\sqrt{|a_n|}}n\leq \frac{|a_n|+\frac 1{n^2}}2.$$ Since the series $\sum_n|a_n|$ and $\sum_n\frac 1{n^2}$ are convergent, we get the convergence of $\sum_n\frac{\sqrt{|a_n|}}n$.
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