Infinite series involving sum of divisors function 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.
 A: 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.
A: 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.
