Absolute convergence of Dyson series for evolution operator In this paper https://arxiv.org/abs/math-ph/9901018 the author (on page 4) mentions : "...a standard argument for the absolute convergence of the Dyson series for [the evolution operator] $\hat{U}_{\tau}$ ..." . 
Could you please give me a hint of what is a standard way of proving the absolute convergence of the Dyson series for the evolution operator? References are welcome.
(I am not a mathematician, but a physicist)
 A: By `Dyson series' I suppose you mean the following expression (with $t\geq t_0$):
$$D (t,t_0) = \sum_{k=0}^\infty \left( - \frac{i}{\hbar} \right)^k 
\int_{\Delta^k(t,t_0)} d^k t \, \prod_{l=1}^k V (t_k)$$
$V\colon t \mapsto V(t)$ is a map on the reals, yielding a linear (unbounded) operator $V(t)$ on your Hilbert space (the domain is the one on which the expectation value of V(t) is finite), and $\Delta^k(t,t_0)$ is the geometric simplex
$$\Delta^k(t,t_0) = \left\lbrace (t_1, \dots, t_k) \in \mathbb{R}^k\middle\vert 
t \geq t_1 \geq t_2 \geq \dots \geq t_k \geq t_0 \right\rbrace . $$
A sufficient and natural condition for convergence of the series is obtained by first defining
$$I \left( t, t_0 \right) := \int_{t_0}^t d t' \, \left\Vert V (t')\right\Vert \, ,$$
and then observing that the last expression of
\begin{align}
\left\Vert D(t,t_0) \right\Vert &\leq \sum_{k=0}^\infty \frac{1}{\hbar^k} \, \left\Vert \int_{\Delta^k(t,t_0)} d^k t \, \prod_{l=1}^k V (t_k) \right\Vert \\ 
& \leq \sum_{k=0}^\infty  \left( \frac{I (t,t_0)}{\hbar} \right)^k
\end{align}
is a geometric series. Thus, we obtain (absolute) convergence for
$$\int_{t_0}^t d t' \, \left\Vert V (t')\right\Vert  \leq \hbar \, .$$
Note that absolute convergence is a must, as else you run into serious problems with defining the limit in the first place.
