Let $\{a_n\}_{n = 0}^\infty$ be a sequence of real numbers such that the series $\sum_\limits{n = 0}^\infty |a_n|^2$ is convergent.

Is it true that $\sum_\limits{n = 0}^\infty \dfrac{a_n}{n}$ is also convergent?

My try:

Since $\sum_\limits{n = 0}^\infty |a_n|^2$ is convergent,

$$\begin{align}\lim_\limits{n\to \infty} |a_n|^2=0 &\implies \forall \epsilon >0 \quad \exists N\in \mathbb N \quad \forall n>N \quad|a_n|^2<\epsilon \\ &\implies \forall n>N \quad |a_n|<\sqrt \epsilon \\ &\implies -\sqrt\epsilon<a_n<\sqrt \epsilon \end{align}$$

Hence $\lim _\limits{n\to \infty} \dfrac{a_n}{n}=0$.

Now $S_n=\sum_\limits{i=1}^n \dfrac{a_i}{n}\implies |S_n|\leq \sum_\limits{i=1}^n \dfrac{|a_i|}{n}\leq \sum _{i=1}^n |a_i|$.

Am I heading in the right way or there is any counter-example? How to complete the proof from here?

Please help.


marked as duplicate by Empty, user147263, Community Oct 12 '15 at 3:43

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.


hint: $\left|\dfrac{a_n}{n}\right| \leq \dfrac{1}{2}\left(a_n^2+\dfrac{1}{n^2}\right)$

  • $\begingroup$ Really appreciate your work $\endgroup$ – Learnmore Oct 12 '15 at 3:43

$$\left(|a_n|-\frac{1}{n}\right)^2\ge 0\implies|a_n|^2-2\left|\frac{a_n}{n}\right|+\frac{1}{n^2}\ge 0$$ $$\implies \left|\frac{a_n}{n}\right|\le \frac{1}{2}\left(|a_n|^2+\frac{1}{n^2}\right)$$


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