# Do asymptotic series at roots of unity uniquely determine a $q$-series?

Let $$\{a_n\}_{n\geq 0}$$ be a sequence of integers. We're interested in specific limits of the formal series $$f(q) = \sum_{n=0}^{\infty}a_n q^n.$$ Let $$\zeta$$ be a root of unity. Say $$\zeta^k=1$$. Formally, when $$q = \zeta e^\hbar$$, $$f(\zeta e^\hbar) = \sum_{n=0}^{\infty}a_n(\zeta e^\hbar)^n = \sum_{j=0}^{\infty}\left(\sum_{n=0}^{\infty}a_n n^j \zeta^n \right)\frac{\hbar^j}{j!} = \sum_{j=0}^{\infty}\left(\sum_{r=0}^{k-1}\zeta^r\sum_{m=0}^{\infty}a_{r+mk} (r+mk)^j\right)\frac{\hbar^j}{j!}.$$ Assume that the sequence $$\{a_n\}$$ is nice enough so that the "$$L$$-series" $$L_{k,r}(s) = \sum_{m=0}^{\infty}\frac{a_{r+mk}}{(m+\frac{r}{k})^s}$$ converges well on $$\text{Re}(s)>M$$ for some $$M \in \mathbb{R}$$ and has analytic continuation as a meromorphic function on $$\mathbb{C}$$ with poles avoiding negative integers and $$0$$.

With this assumption, the following "asymptotic series near $$\zeta$$" is well-defined as a formal power series in $$\hbar$$ : $$f(\zeta e^\hbar) \sim \sum_{j=0}^{\infty}\left(\sum_{r=0}^{k-1}\zeta^r L_{k,r}(-j) k^j\right)\frac{\hbar^j}{j!}$$

Now, the question is, if this asymptotic series is identically $$0$$ near every root of unity $$\zeta$$, does it imply that the original sequence is identically $$0$$?

• The answer in no. An example follows the well known Euler identity $$f(q)=1+\sum_{n\ge 1}(-1)^n\left(q^{\frac{n(3n+1)}{2}}+q^{\frac{n(3n-1)}{2}}\right)=\prod_{n=1}^{\infty}\left(1-q^n\right).$$ Then it is easy to check that $$f(\zeta e^{\hbar})=O_{\zeta}(|\hbar|^A)$$ for any integer $A>0$. That is the corresponding asymptotic series is identically $0$ near every root of unity $\zeta$. – Zhou Sep 28 '19 at 23:03