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 $$ h(q):=(1q)\sum_{n=1}^\infty\frac{q^n}{1q^n}. $$ I can show (with some effort) that $$ h(q)=\log(1q)+f(q) \qquad\text{as}\quad q\to1, \ q<1, $$ 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 $$ h(q)=\log(1q)+c+o(1) \qquad\text{as}\quad q\to1, \ q<1. $$ Then, of course, I am very much interested in the constant $c$. A straightforward computer experiment is not helpful.
Andrew is right, the following limit seems to be what you are looking for $$\lim_{q\uparrow 1}\left(\log(1q)\log q \sum_{n\geq 0}\frac{q^{n+1}}{1q^{n+1}}\right)=\gamma$$ See , for example theorem 1 in "Summations for Basic Hypergeometric Series Involving a $q$Analogue of the Digamma Function" by C. Krattenthaler and H.M. Srivastava. (Though there should be a more canonical source for this somewhere.)

$\begingroup$ Gjergji, this is extremely helpful! (And because your link does not provide a correct link to the paper itself, I add it here: mat.univie.ac.at/~kratt/artikel/digamma.html). I am very pleased to learn that it is actually done by one of my collaborators... $\endgroup$ – Wadim Zudilin Jul 4 '11 at 15:00
The function $h(q)$ is equal to $$\frac{(1q) \left(\log \left(\frac{1}{1q}\right)\psi _q(1)\right)}{\log \left(\frac{1}{q}\right)},$$ where $\psi _q(z)$ is the $q$digamma function. According to Mathematica $с$ is equal to the Euler's constant $\gamma$.

$\begingroup$ Andrew, I definitely vote up, but can only accept Gjergji's answer as it also provides the required details of the evaluation. Mathematica is famous for hiding the tools it actually uses. But I am very thankful to you as well. $\endgroup$ – Wadim Zudilin Jul 4 '11 at 15:03