Assume I have a bound on the summatory function of the form $$\sum_{n \leqslant x} |a_n|^2 \leqslant x^\alpha$$
Can I then deduce something about the convergence of $$\sum \frac{a_n}{n^k} ?$$
I was expecting getting something using Cauchy-Schwarz inequality, but it is not enough since $x^\alpha$ diverge and the sum of $n^{-2k}$ only converges, and does not compensate this $x^a$. Am I missing something?
By the assumption, we have $$ \sum_{n= 2^j}^{2^{j+1}-1 }|a_n|^2 \le C_1 2^{\alpha j} $$ for all $j\ge 0$. Note that $$ \sum_{n= 2^j }^{2^{j+1}-1 }\frac{|a_n|}{n^k}\leq C_22^{-kj}\sum_{n= 2^j }^{2^{j+1} -1}|a_n|\leq C_32^{-(k-\frac{1}{2})j}\sqrt{\sum_{n= 2^j }^{2^{j+1}-1 }|a_n|^2}\le C_42^{-(k-\frac{1}{2}-\frac{\alpha}{2})j} $$ for all $j\ge 0$. Here, $(C_p)_{ 1\le p\le 4}$ is a set of constants depending only on $\alpha$ and $k$. Thus, if $k>\frac{\alpha+1}{2}$, then the sum $$ \sum_{j=0}^\infty\sum_{n= 2^j }^{2^{j+1}-1 }\frac{|a_n|}{n^k} $$ converges absolutely. If $k \le \frac{\alpha+1}{2}$, then $a_n = c\cdot n^{\frac{\alpha-1}{2}}$ satisfies $\sum_{j\leq x} |a_j|^2 \le x^\alpha$ for sufficiently small $c>0$, but $$ \sum_n \frac{a_n}{n^{k}} \ge c\sum_n \frac{1}{n}=\infty. $$ Hence, the bound $k>\frac{\alpha+1}{2}$ cannot be improved.
