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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 $\tfrac{1}{12} + \sum_{n \geq 1} n = 0$.)

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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
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+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
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2 Answers

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.

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+1, Interesting –  Eric Naslund May 23 '11 at 1:33
    
+1, Very Interesting, indeed! –  user02138 May 23 '11 at 1:34
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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 $\frac{1}{1-e^tz} + \frac{1}{e^tz - 1} = 0$.

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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
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