While trying to evaluate a multiple rational zeta series, the following sum came up:
$$\sum_{n=2}^{\infty} \frac{d(n)}{n(n-1)} = \sum_{n=2}^{\infty} \Big{(} \frac{d(n)}{n-1} - \frac{d(n)}{n} \Big{)}. \qquad \qquad \qquad (*) $$ This equality can be obtained by partial fraction decomposition.
I tried evaluating the right-hand side by considering the partial asymptotic sum $$\sum_{n\leq x}\frac{d(n)}{n}=\frac{1}{2}(\log(x))^{2}+2\gamma\log (x)+\gamma^{2}-2\gamma_{1}+O\left(x^{-1/2}\right),$$ and see if there's an analogue for $\sum d(n)/(n-1) $. So far, I haven't succeeded.
Another approach to evaluate $(*)$ is to consider the generating functions
\begin{align} \sum_{n=0}^{\infty} d(n)q^{n} &= \sum_{n=1}^{\infty} \frac{q^{n}}{1-q^{n}} \\ &= \sum_{n=1}^{\infty} \frac{1+q^{n}}{1-q^{n}} q^{n^{2}} \\ &= \frac{1}{(q;q)_{\infty}} \sum_{n=1}^{\infty} (-1)^{n-1} \frac{n q^{\binom{n+1}{2}}}{(q;q)_{n}} \quad, \end{align}
where $(a;q)_{n} := (1-a)(1-aq)(1-aq^{2})\dots(1-aq^{n-1})$ is the $q$-shifted factorial .
One could proceed by dividing either of these expressions by $q^{2}$, integrate twice, and then set $q=1$. However, integrating either of these summands seems like quite a formidable task.
Finally, note that a related sum can be evaluated: $$\sum_{n=1}^{\infty} \frac{d(n)}{n^{2}} = \zeta^{2}(2) = \frac{\pi^{4}}{36} $$ (see e.g. equation (19) here).
Question: can $(*)$ be evaluated? If there aren't any results on this, I'm also interested in other sums that are related to this one of which the exact evaluation is known.
