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 $0$?

  • 1
    $\begingroup$ 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 $\zeta$. $\endgroup$ – Zhou Sep 28 '19 at 23:03

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