MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.)

share|cite|improve this answer
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 $\gamma$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.