We know that the taylor expansion of $$\left(\int_{t}^{t+\Delta t}a(t')dt'\right) = a(t)\Delta t + \frac{1}{2}\frac{da}{dt}(t) \Delta t^2 + \frac{1}{3!}\frac{d^2a}{dt^2}(t) \Delta t^3 + \frac{1}{4!}\frac{d^3a}{dt^3}(t) \Delta t^4 \cdots (1)$$, $a(t)$ can be very big but here I assume that $\Delta t < 1$ so $\Delta t < (\Delta t)^2$.
Are the terms in taylor series always decreasing in higher terms, i.e. $a(t)\Delta t < \frac{1}{2}\frac{da}{dt}(t) \Delta t^2 < \frac{1}{3!}\frac{d^2a}{dt^2}(t) \Delta t^3 < \frac{1}{4!}\frac{d^3a}{dt^3}(t) \Delta t^4$
To well approximate $$\left(\int_{t}^{t+\Delta t}a(t')dt'\right) \approx a(t)\Delta t$$
We need the condition $\frac{1}{2}\frac{da}{dt}(t) \Delta t^2 << a(t) \Delta t$.
But even if the condition holds, the problem may arise even for small but finite $\Delta t$ when extremely large higher derivative occurs:
E.g. $\frac{1}{3!}\frac{d^2a}{dt^2}(t) \Delta t^3 >> \frac{1}{2}\frac{da}{dt}(t) \Delta t^2$ or even $\frac{1}{4!}\frac{d^3a}{dt^3}(t) \Delta t^4 >> \frac{1}{2}\frac{da}{dt}(t) \Delta t^2$. Is $\frac{1}{3!}\frac{d^2a}{dt^2}(t) \Delta t^3 >> \frac{1}{2}\frac{da}{dt}(t) \Delta t^2$ always true for $\Delta t$ small than one? If it is then the radius of convergence is just 1?
What would be its radius of convergence of Eq.(1). Does Theorem of Borel apply in this case and how to make use of it?