The derivative of Riemann zeta function is $\zeta'(s)=-\sum_{n=2}^{\infty}(\log{n}) n^{-s}.$ The square of $\zeta'$ is the following Dirichlet series: $$\zeta'(s)^2=\sum_{n=4}^{\infty}a_nn^{-s},$$ where $a_n=\sum_{k|n}(\log{k})(\log{\frac{n}{k}})$. My question is to estimate the order of the growth of the the coefficient $a_n$ of $(\zeta')^2$.
A trivial estiamtion is given by $a_n\leq d(n)(\log n)^2$. Absolutely, this is not optimal. Do we have a better estimation?
Than you very much!
We have $$ a_n = \log n\sum\limits_{k\mid n} {\log k} - \sum\limits_{k\mid n} {\log ^2 k} \le \log n\sum\limits_{k\mid n} {\log k} - \log ^2 n. $$ Here $$ \sum\limits_{k\mid n} {\log k} = \frac{1}{2}\log\! \Big( {\prod\limits_{k\mid n} {k^2 } } \Big) = \frac{1}{2}\log\! \bigg( {\prod\limits_{k\mid n} {k\frac{n}{k}} } \bigg) = \frac{1}{2}\log n^{d(n)} = \frac{{d(n)}}{2}\log n. $$ Thus, $$ a_n \le \left( {\frac{{d(n)}}{2} - 1} \right)\log ^2 n. $$