# Laurent Series $\sum_{n=-1}^\infty a_n x^n$ when $a_{-1} = \infty$

When dealing with complex functions, if $$f(x)$$ has a simple pole, then we can find the coefficient $$a_{-1}$$ in the Laurent expansion $$f(x) = \sum_{n=-1}^\infty a_n x^n$$ by evaluating the limit $$\lim_{x \to 0} f(x)x$$.

I'm interested in an odd sort of edge case, where we force $$a_{-1} = \infty$$. For instance, consider the function $$f(x) = \sum_{n=1}^\infty n \frac{x^n}{1-x^n}$$ Each of these terms has a simple pole at $$x=1$$. So, if we tried to find the coefficient of the simple pole, we would obtain $$\sum_{n=1}^\infty n \frac{x^n}{1-x^n} = \sum_{n=1}^\infty n \left(-\frac{1}{n}\frac{1}{x-1} + \sum_{n=0}^\infty b_n (x-1)^n\right) = \left(-\sum_{n=1}^\infty 1\right)\frac{1}{1-x} + \sum_{n=0}^\infty c_n (x-1)^n$$

But this computation suggests $$a_{-1} = - \sum_{n=1}^\infty 1$$, which is infinite. And indeed, we do find that $$\lim_{x \to 1} f(x)(x-1) = \infty$$. However, to my surprise, $$f(x)$$ appears to actually have a second order pole at $$x = 1$$. In fact, computations suggest that $$\lim_{x \to 1} f(x)(x-1)^2 = \zeta(2)$$. So, it seems that somehow, the fact that $$a_{-1}$$ is infinite causes it to spill over to the next order pole.

This behavior continues for other powers of $$n$$, for instance, $$\sum_{n=1}^\infty n^2 \frac{x^n}{1-x^n} = \left(-\sum_{n=1}^\infty n\right)\frac{1}{1-x} + \sum_{n=0}^\infty c_n (x-1)^n$$
And then $$\sum_{n=1}^\infty n$$ informally has a size of about $$\infty^2$$, so it should spill over to the next two poles. And $$f(x)$$ has a third order pole with $$\lim_{x \to 1} f(x)(x-1)^3 = 2! \zeta(3)$$. Likewise plugging in $$n^3$$ gives a fourth-order pole with $$\lim_{x \to 1} f(x)(x-1)^4 = 3! \zeta(4)$$.

An interesting case is $$g(x) =\sum_{n=1}^\infty \frac{x^n}{1-x^n}$$, since it should have a pole like $$\left(\sum_{n=1}^\infty \frac{1}{n}\right) \frac{1}{1-x}$$ $$\sum_{n=1}^\infty \frac{1}{n} \approx \ln(\infty)$$, so we should expect $$g(x)$$ to have a pole of order $$\frac{\ln(1-x)}{(1-x)}$$, and computationally, we see to have $$\lim_{x \to 1} g(x) \frac{(1-x)}{\ln(1-x)} \approx 1$$.

## Question

A natural conjecture to make is that the leading term of the asymptotic of $$\sum \frac{a_n}{n}$$ determines the type of pole $$f(x) = \sum_{n=1}^\infty a_n \frac{x^n}{1-x^n}$$ has. In particular, if the leading term of $$\sum \frac{a_n}{n} = g(n)$$, then does the dominant pole of $$f(x)$$ have order $$\frac{g\left(\frac{1}{1-x}\right)}{(1-x)}$$?

I'm also curious about computing $$\lim_{x \to 1} f(x) \frac{g\left(\frac{1}{1-x}\right)}{(1-x)}$$. For instance, is there a way to establish the result that $$\lim_{x \to 1} \left((1-x)^k \sum_{n=1}^\infty n^k \frac{x^n}{1-x^n}\right) = (k-1)! \zeta(k)$$ Is there a systematic way to evaluate these limits in general?

• $$\sum_{n\ge1}n^kx^n/(1-x^n)=\sum_{n\ge1}\sigma_k(n)x^n$$ where $\sigma_k(n)$ is the sum of the $k$th powers of the divisors of $n$. So when $k$ is odd this is the Eisenstein series $E_{k+1}$ up to scaling and constant term. When $k$ is even, this has also been studied, by Ingham I believe. Apr 24, 2023 at 17:02
• These functions have singularities at all $n$th roots of $1$, so $1$ is not an isolated singularity and in particular not a pole. Apr 24, 2023 at 18:07

There is an asymptotic for sums of the form $$g(t) = \sum_{n=1}^{\infty} f(nt)$$ given by Zagier for $$C^{\infty}$$ functions $$f$$ of sufficiently rapid decay (Proposition 3 of his appendix on the Mellin transform, https://people.mpim-bonn.mpg.de/zagier/files/tex/MellinTransform/fulltext.pdf)
If $$f(t) \sim \sum_{n=0}^{\infty} b_n t^n, \quad (t \rightarrow 0)$$ then $$g(t) \sim \frac{1}{t} \int_0^{\infty} f(t) \, \mathrm{d}t + \sum_{n=0}^{\infty} b_n \frac{B_{n+1}}{n+1} (-t)^n.$$
The example of $$\sum_{n=1}^{\infty} n^{k-1} \frac{e^{-tn}}{1 - e^{-tn}}$$ is given there specifically, as Example 3. We have the asymptotic series $$\sum_{n=1}^{\infty} n^{k-1} \frac{e^{-tn}}{1 - e^{-tn}} \sim \frac{(k-1)! \zeta(k)}{t^k} + \sum_{r=0}^{\infty} (-1)^{r+k} \frac{B_r B_{r+k-1}}{r! (r+k-1)!}t^{r-1}$$ as $$t \rightarrow 0$$, and in particular with $$x = e^{-t}$$ $$\lim_{x \rightarrow 1} (1-x)^k \sum_{n=1}^{\infty} n^{k-1} \frac{x^n}{1-x^n} = \lim_{t \rightarrow 0} t^k \sum_{n=1}^{\infty} n^{k-1} \frac{e^{-tn}}{1 - e^{-tn}} = (k-1)! \zeta(k).$$
For real $$k>0$$, let $$$$L_k:= \lim_{x\uparrow1}\Big((1-x)^{k+1} \sum_{n=1}^\infty n^k \frac{x^n}{1-x^n}\Big) =\lim_{t\downarrow0}\sum_{n=1}^\infty a_n(t),$$$$ where $$$$a_n(t):=t^{k+1}n^k \frac{(1-t)^n}{1-(1-t)^n}.$$$$ For $$t\in(0,1)$$ and natural $$n$$,
$$$$a_n(t)\le b_n(t):=t^{k+1}n^k \frac{e^{-nt}}{1-e^{-nt}} =t\,\frac{(nt)^k}{e^{nt}-1}$$$$ and $$$$a_n(t)\sim b_n(t)$$$$ if $$nt\asymp1$$. Also, for $$t\in(0,1)$$ and natural $$n$$,
$$$$b_n(t)=\int_{nt}^{(n+1)t}dc\,\frac{(nt)^k}{e^{nt}-1} \le4\int_{nt}^{(n+1)t}dc\,\frac{c^k}{e^c-1}$$$$ and $$$$b_n(t)\sim \int_{nt}^{(n+1)t}dc\,\frac{c^k}{e^c-1}$$$$ if $$nt\asymp1$$. It follows that $$$$L_k=\lim_{h\downarrow0}\int_h^{1/h}dc\,\frac{c^k}{e^c-1} =\int_0^\infty dc\,\frac{c^k}{e^c-1}=\Gamma(1+k) \zeta(1 + k).$$$$