Is there much known about infinite series of the form
$\sum_{n=1}^{\infty}\frac{\sigma_{1}(n)}{n}q^{n}$
where $\sigma_{1}(n)$ is the sum of divisors function.
I am particularly interested in finding a solution to the above series with (1/2) replacing q.
On
I would like to point out that there is an approximation that is good to an amazing $24$ digits of the value of the sum for $q=1/2$ that can be obtained using harmonic summation techniques.
Introduce the sum $$S(x) = \sum_{n\ge 1} \frac{1}{n}\frac{1}{2^{nx}-1}$$ so that we are interested in $S(1).$
The sum term is harmonic and may be evaluated by inverting its Mellin transform.
Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have $$\lambda_k = \frac{1}{k}, \quad \mu_k = k \quad \text{and} \quad g(x) = \frac{1}{2^x-1}.$$
We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty \frac{2^{-x}}{1-2^{-x}} x^{s-1} dx = \int_0^\infty \sum_{q\ge 1} e^{-(\log 2)q x} x^{s-1} dx \\= \Gamma(s) \frac{1}{(\log 2)^s} \sum_{q\ge 1} \frac{1}{q^s} = \frac{1}{(\log 2)^s} \Gamma(s) \zeta(s).$$
Since $1/(2^x-1)\sim 1/x/\log(2)$ in a neighborhood of zero and $1/(2^x-1)\sim 2^{-x}$ at infinity we have that the fundamental strip of this transform is $\langle 1, \infty\rangle.$
It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x)$ is given by
$$Q(s) = \frac{1}{(\log 2)^s} \Gamma(s) \zeta(s) \zeta(s+1) \\ \text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} \frac{1}{k} \frac{1}{k^s} = \sum_{k\ge 1} \frac{1}{k^{s+1}} = \zeta(s+1)$$ for $\Re(s) > 0.$
The Mellin inversion integral here is $$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the left for an expansion about zero where the line is chosen in the intersection of the fundamental strip of the transform with the half-plane of convergence of the harmonic factor that multiplies $g^*(s).$
We now compute the inversion integral. Fortunately this calculation is very simple since the trivial zeros of the two zeta function terms together cancel the poles of the gamma function term. What remains is just three residues.
They are: $$\mathrm{Res}(Q(s)/x^s; s=1) = \frac{\pi^2}{6} \frac{1}{x \log 2},$$ $$\mathrm{Res}(Q(s)/x^s; s=0) = -\frac{1}{2}\log(2\pi) + \frac{1}{2} \log\log 2 + \frac{1}{2} \log x$$ and finally $$\mathrm{Res}(Q(s)/x^s; s=-1) = -\frac{1}{24} x \log 2.$$
Setting $x=1$ we obtain the following approximation of $S(1):$ $$\frac{\pi^2}{6\log 2} -\frac{1}{2}\log(2\pi) + \frac{1}{2} \log\log 2 -\frac{1}{24}\log 2$$ which is $$\frac{\pi^2}{6\log 2} -\frac{1}{2}\log\pi + \frac{1}{2} \log\log 2 -\frac{13}{24}\log 2.$$
This gives the value $$1.24206209481241494579784529798$$ while the exact value is $$1.24206209481241494579784548189$$ so the approximation is good to $24$ digits the difference being $$-{ 1.83904\times 10^{-25}}.$$
This MSE link contains a calculation in the same spirit, but somewhat more advanced.
Note that $\frac{\sigma_0(n)}{n}=\sum_{d\mid n} \frac{1}{d}$.
Then
$$\begin{align}f(q)&=\sum_{n} \frac{\sigma_0(n)}{n}q^n\\& = \sum_n\sum_{d\mid n} \frac{1}{d}q^n \\ &=\sum_{d=1}^\infty \frac{1}{d}\sum_{m=1}^\infty q^{md} \\&=\sum_{d=1}^\infty \frac{1}d\frac{q^d}{1-q^d} \end{align}$$
Not sure if that helps. In particular, for $q=1/2$ this is:
$$\sum_{d=1}^\infty \frac{1}{d(2^d-1)}$$
Now, $\sum_{d=1}^\infty \frac{1}{d}q^d = \log \frac{1}{1-q}$.
So $$ f(q)-\log{\frac{1}{1-q}} = \sum_{d} \frac{1}{d}\frac{q^{2d}}{1-q^d}$$
By induction, we see that:
$$f(q)-\sum_{k=1}^n \log{\frac{1}{1-q^{k}}} = \sum_d \frac{1}{d}\frac{q^{(n+1)d}}{1-q^d}$$
So we can see the limit as $n\to\infty$ is $0$ for $|q|<1$. So:
$$f(q) = \log \prod_{k=1}^\infty \frac{1}{1-q^k}$$
The function $\prod_k \frac{1}{1-q^k}$ is the generating function for the partition function, so $f(q)$ is the logarithm of that function. It is unlikely to be able to simplify this, since the generating function for partitions is hard to simplify.