# Is there a "quantum" Riemann zeta function?

Occasionally I find myself in a situation where a naive, non-rigorous computation leads me to a divergent sum, like $\sum_{n=1}^\infty n$. In times like these, a standard approach is to guess the right answer by assuming that secretly my non-rigorous manipulations were really manipulating the Riemann zeta function $\zeta(s) = \sum_{n=1}^\infty n^{-s}$ and its cousins. Then it's reasonable to guess that the "correct" answer is, for example, $\sum_{n=1}^\infty n = \zeta(-1) = -\frac1{12}$. Thus the zeta function and its cousins are a valuable tool for other non-number-theoretic problem solving: it's always easier to rigorously prove that your guess is correct (or discover, in trying to prove it, that it's wrong) than it is to rigorously derive an answer from scratch.

I recently found myself wishing I could do something similar for the sum of the quantum integers. Recall that at quantum parameter $q = e^{i\hbar}$, quantum $n$ is the complex number $$[n]_q = \frac{q^n - q^{-n}}{q - q^{-1}} = q^{n-1} + q^{n-3} + \dots + q^{3-n} + q^{1-n}.$$ The point is that $[n]_1 = n$.

Question: Are there established methods to sum the divergent series $\sum_{n=1}^\infty [n]_q$ and its cousins? For example, is there some well-behaved function $\zeta_q(s)$ for which the series is naturally the $s=-1$ value?

Note that when $n$ is a root of unity, the series truncates, and it would be nice (but maybe too much too hope for) if the regularized series agreed with the truncated series at these values.

I should mention also that I consider the following answer tempting but inaccurate, as it definitely doesn't work at roots of unity, which I do care about:

$$\sum_{n=1}^\infty [n]_q = \frac1{q-q^{-1}} \sum_{n=1}^\infty (q^n - q^{-n}) = \frac1{q-q^{-1}} \left( \sum_{n=1}^\infty q^n - \sum_{n=1}^\infty q^{-n}\right) =$$ $$= \frac1{q-q^{-1}} \left( \frac{q}{1-q} - \frac{q^{-1}}{1-q^{-1}}\right) = \frac{q+1}{(q-q^{-1})(q-1)}$$

• $= \frac{q}{(1-q)^2}$ Commented Nov 24, 2010 at 4:36
• Seems like you missed a negative sign at the last equality.:) And the way you did it is like the way Euler find $\sum n$ is -1/12 Commented Nov 30, 2010 at 2:32
• @Vivek: Yes, I didn't simplify it enough. @Yuhao: quite likely --- I wasn't that careful, because I was hoping for an answer better than what I can do. Commented Nov 30, 2010 at 3:24

The paper by Cherednik On q-analogues of Riemann's zeta function (arXiv:math/9804099) gives precisely the definition you're after: $$\zeta_q(s)=\sum\limits_{n=1}^\infty q^{sn}/[n]_q^s$$ His paper also contains a brief discussion of the properties of this $$q$$-zeta function. On the other hand, the term quantum zeta function appears to have a somewhat different meaning, see e.g. the paper On the quantum zeta function by R.E. Crandall.
One key point is that the value of the function $$\zeta_q$$ at negative integers is a fraction which has no limit when $$q$$ goes to $$1$$. One can obtain a relation to the $$q$$-Bernoulli numbers introduced by Carlitz in 1948, by taking a difference with the value of a modified $$\zeta_q$$ function.