The following (very simply looking!) problem occurs in regularization
of the harmonic series
which can be formally thought of as the limit as $q\to1$, $|q|<1$, of
$$
h(q):=(1-q)\sum\_{n=1}^\infty\frac{q^n}{1-q^n}.
$$
I can show (with some effort) that
$$
h(q)=-\log(1-q)+f(q) \qquad\text{as}\quad q\to1, \ |q|<1,
$$
where $f(q)$ is a bounded function (hint: consider both $h(q)$
and $h(q^2)$ as $q\to1$). The question is whether the function
$f(q)$ has a limit as $q\to1$ or not; in other words, whether
$$
h(q)=-\log(1-q)+c+o(1) \qquad\text{as}\quad q\to1, \ |q|<1.
$$
Then, of course, I am very much interested in the constant $c$.
A straightforward computer experiment is not helpful.



Asymptotics of the $q$-harmonic series as $q\to1$
Andrew is right, the following limit seems to be what you are looking for
$$\lim\_{q\uparrow 1}\left(\log(1-q)-\log q \sum\_{n\geq 0}\frac{q^{n+1}}{1-q^{n+1}}\right)=\gamma$$
See , for example theorem 1 in ["Summations for Basic Hypergeometric Series Involving a $q$-Analogue of the Digamma Function"](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.25.214) by C. Krattenthaler and H.M. Srivastava. (Though there should be a more canonical source for this somewhere.)


