Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Dear Math Stackexchange,

I'm a physics researcher working on a problem in quantum statistical physics. I've encountered the following function which I do not recognize (neither does Mathematica):

$$R(z; a,b) =\sum_{k=0}^{\infty} \frac{a^k}{1+z \, b^k},$$

where $a$ and $b$ are real parameters, $b>1$ and $0<a<b$.

In practice I need values of this function for $z$ on the real positive semi-axis, but I would like to understand it better first. As far as I see, $R(z; a,b)$ is a meromorphic function with poles within the unit circle and an essential singularity at $z=0$.

Can it be related to some known special function? Are there more efficient ways to compute it rather than to sum the defining series directly? I'd appreciate any advice or hint. Physical context for this problem can be found on Physics SE.

share|cite|improve this question
up vote 7 down vote accepted

At least when $|z|>1$, $$ R(z;a,b)=\sum_{k=0}^{+\infty}(-1)^k\frac{b^{k+1}}{b^{k+1}-a}\left(\frac1z\right)^{k+1}. $$ Thus, $R(\,\cdot\,;a,b)$ may be expressed as a $q$-hypergeometric function ${}_2\phi_1$. In the special case when $a=1$, $R(z;1,b)=-L_1(-b/z,1/b)$ where $L_1(\,\cdot\, ;q)$ is the $q$-logarithm.

share|cite|improve this answer
Thanks! This seems to be the way to go (numerically) outside the unit circle, although approaching the $|z|=1$ the convergence naturally slows down. – Slaviks Aug 20 '11 at 13:56
Yes. With the exception that if $z$ is real and $1\le z\le b$ (the case $z=1$ included) then for every $a$ the series is alternated. Hence $R(z)$ is between two successive partial sums. Don't know if this helps. – Did Aug 20 '11 at 14:07
For searching purposes: ${}_p \phi_q$ is sometimes referred to as a basic hypergeometric series... – J. M. Aug 21 '11 at 16:14
Still struggling with $|z|<1$ though. It is easy to prove that $R(z; a,b) = \frac{1}{1-a} - R(1/z; a,1/b)$ but I have difficulty to link it to a convergent $q$-series. – Slaviks Aug 22 '11 at 13:07

Completing Didier's solution, the answer for $|z|>1$ in terms of $q$-hypergeometric function is

$$R(z; a,b) = \frac{1}{a-1} \left ( {}_2\phi_1\left [ \begin{matrix} 1/b , \, a \\ a/b \end{matrix} ; 1/b, -1/z \right ] -1 \right ) .$$

share|cite|improve this answer

After getting used a bit to $q$-hypergeometric function , I found a way to re-arrange the series for $|a|<1$ and $z>0$:

If $b>1$, $$R(z; a,b) =\frac{1}{1-a}- \sum_{k=0}^{\infty} \frac{a^k}{1+b^{-k} z^{-1}}=\frac{1}{1-a}- {}_2\phi_1\left [ \begin{matrix} -1/z , \, 1/b \\ -1/(b z) \end{matrix} ; 1/b, a \right ] \frac{z}{1+z} .$$

If $b<1$, $$R(z; a,b) = {}_2\phi_1\left [ \begin{matrix} -z , \, b \\ -b \, z \end{matrix} ; b, a \right ] \frac{1}{1+z} .$$

This covers all the relevant cases for the original physics problem.

share|cite|improve this answer
I haven't taken a good look at basic hypergeometric series, but the usual hypergeometric series allows for transformations like these, which I presume are just the limiting cases as $q\to 1^-$. – J. M. Aug 23 '11 at 3:49

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.