If $\sum (a_j)^2$ abs converge , does $\sum a_j / (j+1) $ that too?

Question

Suppose that $\sum (a_j)^2$ converge absolute, how can I prove that $\sum a_j / (j+1)$ does that too? I don't exactly know where to start to prove this. Can somebody help me or give a hint?

Answer 1

Hint. Note that by Cauchy-Schwarz inequality, for any positive integer $n$, $$\left(\sum_{j=1}^n \frac{|a_j|}{j+1}\right)^2\leq \sum_{j=1}^n a_j^2\cdot\sum_{j=1}^n \frac{1}{(j+1)^2}.$$