Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\Phi_{l}(z;q)$ denote the Lerch Transcendent and $Li_{l}(z)$ denote the Polylogarithm function. For $l \in \mathbb{Z}_{\geq 0}$, we have the following Laurent expansion \begin{align} \Phi_{-l}(z;0) + (-1)^{l} Li_{-l}(1/z) = \sum_{k \geq 0} k^{l} z^{k} + \sum_{k \leq -1} k^{l} z^{k}. \end{align} For $l \in \mathbb{Z}_{\geq 0}$, the left side will always be a sum of two rational functions which cancel. For example, $l = 0$ and $l = 1$ gives \begin{align} \frac{1}{1-z} + \frac{1}{(1-1/z)z} = 0 \quad \text{and} \quad \frac{z}{1-z} - \frac{1}{(1-1/z)^{2} z} = 0. \end{align} Looking at it another way, I wonder if there is some simple reason why the right side is zero, which can be written (as the meaningless sum) $$ \sum_{k \in \mathbb{Z}} k^{l} z^{k} = 0 \quad \text{for}\quad l \in \mathbb{Z}_{\geq 0}.$$ Is there some way to interpret this expression and show that its value is zero?

(Compare this to the meaningless expression $\tfrac{1}{12} + \sum_{n \geq 1} n = 0$.)

share|improve this question
How did you conclude the rational functions on the left cancel? –  Eric Naslund May 23 '11 at 0:26
A problem I am having is that the right hand side does not converge for any $z\neq 0$ since the terms in one direction have their size going to infinity. Because of this it seems like a meaningless expression? –  Eric Naslund May 23 '11 at 0:31
+1, admittedly, (based on the great answers!) there is quite an interesting question here! I personally just hit a bit of a "mental roadblock" when I see that series do not converge... –  Eric Naslund May 23 '11 at 15:13

2 Answers 2

up vote 5 down vote accepted

As several people have already points out, the actual sum $\sum_{n=-\infty}^{\infty} n^k z^n$ is nowhere convergent. However, there is an interesting phenomenon here.

Let $M$ be the group of functions $\phi: \mathbb{Z}^k \to \mathbb{Q}$ where $\mathbb{Z}^k$ can be partitioned into finitely many lattice polytopes, such that $\phi$ is polynomial on each piece.

Theorem (Lawrence): There is a unique linear map $h: M \to \mathbb{Q}(z_1, \ldots, z_k)$ such that, if $\sum \phi(a_1, \ldots, a_k) z_1^{a_1} \cdots z_k^{a_k}$ converges anywhere, then $h(\phi)$ is equal to this sum.

For example, let $\phi : \mathbb{Z} \to \mathbb{Q}$ be $n \mapsto |n|$. Then $\sum \phi(n) z^n$ is nowhere convergent. However, $\phi(n) = \phi_1(n) + \phi_2(n)$, where $\phi_1(n)$ is $n$ if $n > 0$ and $0$ otherwise, and $\phi_2(n)$ is $-n$ if $n<0$ and $0$ otherwise. The sums $\sum \phi_1(n) z^n$ and $\sum \phi_2(n) z^n$ converge to $z/(1-z)^2$ and $z^{-1} /(1-z^{-1})^2$ respectfully, so $h(\phi) = h(\phi_1 + \phi_2) = z/(1-z)^2 + z^{-1} /(1-z^{-1})^2$. The theorem of Lawrence is a precise statement of the fact that one can work with such expressions consistently.

Moreover, define $x_i \phi$ to be the function $(a_1, \ldots, a_n) \mapsto \phi(a_1, \ldots, a_i -1, \ldots, a_n)$. Then one can check that $h(x_i \phi) = x_i \phi$. This formula is obvious whenever the defining sum converges, and it is not too bad to see that it must then hold for all $\phi$.

Once you believe this, there is a nice explanation for your observation. You are interested in $h(n \mapsto n^k)$. We have $$(1-x_1)^{k+1} h(n \mapsto n^k) = h \left(n \mapsto n^k - k (n-1)^k + \binom{k}{2} (n-2)^k - \cdots \pm (n-k-1)^k \right) = h(0) = 0$$

The middle equality is using the identity that the $k+1$-st difference of a degree $k$ polynomial is $0$. For example, $n^2 - 3 (n-1)^2 + 3 (n-2)^2 - (n-3)^2 =0$.

Now, $(1-x_1)^{k+1}$ is not a zero divisor in $\mathbb{Q}(x_1)$. So we conclude that $h(n \mapsto n^k)=0$, as you observed.

As far as I know, Lawrence's paper is not available online. The reference is Jim Lawrence, Rational-function-valued valuations on polyhedra, in Discrete and computational geometry (New Brunswick, NJ, 1989/1990), 199–208, DIMACS Ser. Discrete Math. Theoret. Comput. Sci. 6, Amer. Math. Soc. (1991). There is also some good discussion of this in Barvinok's lectures.

share|improve this answer
+1, Interesting –  Eric Naslund May 23 '11 at 1:33
+1, Very Interesting, indeed! –  user02138 May 23 '11 at 1:34

For nonnegative integers $\ell$ we have

$\Phi_{-\ell}(z;0) = \sum_{n=0}^\infty n^\ell z^n$

converging for $|z| < 1$ while

$(-1)^\ell {\rm Li}_{-\ell}(1/z) = \sum_{n=-\infty}^{-1} n^\ell z^n$

converging for $|z|>1$. The combined Laurent series $\sum_{n=-\infty}^\infty n^\ell z^n$ diverges everywhere, so this part doesn't make sense. However, as for the rational functions

$\Phi_{-\ell}(z;0) = ((\frac{\partial}{\partial t})^\ell \frac{1}{1- e^{t} z})|_{t=0} $

$(-1)^\ell {\rm Li}_{-\ell}(1/z) = ((\frac{\partial}{\partial t})^\ell \frac{1}{e^t z - 1})|_{t=0}$

and of course $\frac{1}{1-e^tz} + \frac{1}{e^tz - 1} = 0$.

share|improve this answer
Yes, this is what I had in mind! –  user02138 May 23 '11 at 1:10
Is there a similar expression for the rational function $\Phi_{-l}(z,q)$ for integer $q$? –  user02138 May 23 '11 at 1:31
$\Phi_{-\ell}(z;q) = \sum_{n=0}^\infty (n+q)^\ell z^n = ((\frac{\partial}{\partial t})^\ell \frac{e^{q t}}{1 - e^t z} ) |_{t=0}$ –  Robert Israel May 23 '11 at 7:07
Thanks! –  user02138 May 23 '11 at 7:24

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.