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)

• Please don't let us do all the work: What is the operator $\hat{U}_\tau$, so at least give us the Hamiltonian, then it is possible to say more. – Luke May 23 '17 at 12:33
• Thanks for your reply. The paper just says that the Hamiltonian is a self-adjoint, bounded operator. – Salsiccio May 23 '17 at 13:49

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 (say bounded) operator $$V(t)$$ on your Hilbert space, 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} D(t,t_0) &\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.