# If $\sum\limits_{n=1}^∞u_n^2$ is convergent, then $\sum\limits_{n=1}^∞\frac{u_n}n$ is absolutely convergent [duplicate]

If $\{u_n\}$ is a sequence of real numbers and the series $\displaystyle\sum_{n=1}^{\infty}u_n^2$ is convergent, prove that the series $\displaystyle\sum_{n=1}^{\infty}\frac{u_n}{n}$ is absolutely convergent.

I have proved the series $\displaystyle\sum_{n=1}^{\infty}\frac{u_n}{n}$ is convergent, but how to prove it is absolutely convergent?

## marked as duplicate by Martin R, Gabriel Romon, Math1000, Community♦May 16 '18 at 10:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• I use latex math flashcards in android – Auro Das May 16 '18 at 8:55
• This must be the 20-th time this question is asked here ... – Gabriel Romon May 16 '18 at 9:01
• – Martin R May 16 '18 at 9:07
• Btw, if you have already proven that $\sum_{n=1}^{\infty}(\frac{u_n}{n})$ is convergent then you are done: Just apply the same reasoning to $(|u_n|)$. – Martin R May 16 '18 at 9:18

## 2 Answers

One may use$$2\cdot \frac{|u_n|}n \le |u_n|^2+\frac{1}{n^2},\quad n\ge1.$$

• (proof?)....... – James May 16 '18 at 8:56
• @JimmySabater We have$$0\le\left( |u_n|-\frac1n\right)^2$$ – Olivier Oloa May 16 '18 at 8:57

Apply Cauchy-Schwarz:$$\sum_{i=1}^N\left|\frac{u_n}n\right|\leqslant\sqrt{\sum_{i=1}^n{u_i}^2}\sqrt{\sum_{i=1}^n\frac1{i^2}}.$$