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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:

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

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$ = \frac{q}{(1-q)^2} $ –  Vivek Shende Nov 24 '10 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 –  Yuhao Huang Nov 30 '10 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. –  Theo Johnson-Freyd Nov 30 '10 at 3:24
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2 Answers

up vote 27 down vote accepted

The paper by Cherednik On q-analogues of Riemann's zeta function 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.

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Here is another article dealing with similar functions:

q-analogue of Riemann’s ζ-function and q-Euler numbers. by Junya Satoh.

There are also many articles by Taekyun Kim on related functions.

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.

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