We can determine the asymptotics of partial sums involving the divisor function accurately by means of, for example, the hyperbola method: $$\sum_{n\leq N}\frac{d(n)}{n}=\frac{1}{2}(\log(N))^{2}+2\gamma\log (N)+\gamma^{2}-2\gamma_{1}+O\left(N^{-1/2}\right). \qquad \qquad (*)$$ A derivation of the result above by means of this method can be found in Eric Naslund's answer over here. The expression can also be obtained through the use of the Mellin-Perron summation formula.
I wonder whether the result may also be found through the use of generating functions. Recall that: $$ \sum_{n=0}^{\infty} d(n)q^{n} = \sum_{n=1}^{\infty} \frac{q^{n}}{1-q^{n}} .$$ If we slightly adapt this result to our purposes by means of this answer, we find that $$\sum_{n=1}^{N} d(n)q^{n} = \sum_{n=1}^{\infty} \frac{q^{n(N+1)}}{q^{n}-1} - \frac{q^{n}}{q^{n}-1} .$$
So if we now divide by $q$, integrate from $q=0$ to $q=z$, switch integral and summation operators, and let $z \to 1$, we obtain an expression for the titular sum. Let's integrate term by term:
$$\int \frac{q^{n-1}}{q^{n}-1} dq = \frac{\log(1-q^{n})}{n} + C_{1} ,$$ and $$\int \frac{q^{n(N+1)-1}}{q^{n}-1} dq = - q^{n(N+1)} \frac{\ {_2F_1}\left(1,N+1;N+2;q^{n}\right)}{n(N+1)} + C_{2} .$$ Here, the later equality involves a hypergeometric series.
So these expressions need to be evaluated in the aforementioned integration limits and summed over $n \in \mathbb{Z}_{\geq 1}$. Now, I'm not sure how to continue.
Question: can an asymptotic expression akin to the one in $(*)$ - i.e., one that includes the constant terms - be obtained by pursuing this generating functions method further?
