Is any closed-form representation known for the sum $\sum\limits_{n=1}^{\infty}\frac{\mu(n)\log n}{n^2}$?

Solution 1:

You can use a series for the reciprocal Riemann $\zeta$-function: $\zeta(s)^{-1}=\sum\limits_{n=1}^{\infty}\mu(n)n^{-s}$, then take a derivative with respect to $s$ and let $s=2$. This gives you $\sum\limits_{n=1}^{\infty}\frac{\mu(n)\log n}{n^2}=\frac{\zeta'(2)}{\zeta(2)^2}$. It is well-known that $\zeta(2)=\frac{\pi^2}{6}$, and a closed form of the derivative is given in OEIS A073002:

$\zeta'(2)=\frac{\pi^2}{6}\left(\gamma+\log(2\pi)-12 \log A\right)$,

where $\gamma$ is the Euler-Mascheroni constant, and $A$ is the Glaisher-Kinkelin constant. Taking it all together, the result is:

$\sum\limits_{n=1}^{\infty}\frac{\mu(n)\log n}{n^2}=\frac{6}{\pi^2}\left(\gamma+\log(2\pi)-12 \log A\right)$.