I am reading the proof that theta series are holomorphic in John Voight's Quaternion Algebras (Lemma 40.4.2), and I've run into an issue as my analysis is quite rusty (or perhaps was never that strong to begin with...).
Let $Q(x)$ be a positive definite quadratic form in $m = 2k$ variables and $\theta(z) = \sum_{n=0}^\infty r_Q(n) q^n$, where $r_Q(n)$ is the representation number of $n$ by $Q$. The author gives the following quick argument:
- Since $Q$ is positive definite, there is $c \in \mathbb{R}_{\geq 0}$ such that $Q(x) \geq c(x_1^2 + \cdots + x_m^2)$.
- Thus $r_Q(n) = O(n^k)$, and $\theta(z) \leq M \sum_{n=0}^\infty n^k q^n$.
- Thus $\theta(z)$ is holomorphic.
I'm trying to fill in the details for $2$ implies $3$. I assume this is some simple application of the $M$-test, but I can't quite see it. Honestly, it's not even entirely clear to me that the $q$-series bounding $\theta(z)$ is holomorphic itself...
Assuming the author is using the convention that $q=e^{2\pi iz}$, so for every $\delta>0$, if $\Im(z)\ge\delta$, then
$$ |\theta(z)|\le1+\sum_{n\ge1}r_Q(n)e^{-2\pi n\delta}<1+M\sum_{n\ge1}n^ke^{-2\pi n\delta}, $$
so by the Weierstrass M-test, the $q$-series of $\theta(z)$ converges uniformly in every compact subset of the upper half plane $\mathbb H=\{z\in\mathbb C:\Im(z)>0\}$. This indicates that $\theta(z)$ is a holomorphic function in $\mathbb H$.