Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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.

share|cite|improve this question
See here for a proof. – fermesomme Jul 1 '14 at 8:40
up vote 18 down vote accepted

The Cauchy-Schwarz inequality gives $$\sum_{n=1}^\infty \frac{\sqrt{a_n}}{n}\leq \sqrt{\sum_{n=1}^\infty a_n}\,\sqrt{\sum_{n=1}^\infty \frac{1}{n^2}}<\infty.$$

share|cite|improve this answer

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 $\sum_n\frac{\sqrt{|a_n|}}n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.