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
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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?



Laurent Series $\sum_{n=-1}^\infty a_n x^n$ when $a_{-1} = \infty$
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 $x = e^{-t}$ $$\lim\_{x \rightarrow 1} (1-x)^k \sum\_{n=1}^{\infty} n^{k-1} \frac{x^n}{1-x^n} = \lim\_{t \rightarrow 0} t^k \sum\_{n=1}^{\infty} n^{k-1} \frac{e^{-tn}}{1 - e^{-tn}} = (k-1)! \zeta(k).$$


