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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.

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up vote 9 down vote accepted

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" by C. Krattenthaler and H.M. Srivastava. (Though there should be a more canonical source for this somewhere.)

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Gjergji, this is extremely helpful! (And because your link does not provide a correct link to the paper itself, I add it here: I am very pleased to learn that it is actually done by one of my collaborators... – Wadim Zudilin Jul 4 '11 at 15:00

The function $h(q)$ is equal to $$\frac{(1-q) \left(\log \left(\frac{1}{1-q}\right)-\psi _q(1)\right)}{\log \left(\frac{1}{q}\right)},$$ where $\psi _q(z)$ is the $q$-digamma function. According to Mathematica $с$ is equal to the Euler's constant $\gamma$.

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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. – Wadim Zudilin Jul 4 '11 at 15:03

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