I have shown that $F(x)=\int_0^\infty\frac{e^{-t}}{(1+xt)^2}dt\sim\sum_{n=0}^\infty (-1)^n (n+1)! x^n$ as $x\rightarrow 0_+$.
My question now is that when we are given a small value $x$ how can I find the value of $n$ in terms of smallest magnitude? For example what is the exact value of $F(0.1)$ by optimal truncation?
The function $f(n) = (n+1)! x^n$ is convex, so the value of $n$ which minimizes $f(n)$ is the smallest $n$ for which
$$ f(n) \leq f(n+1). $$
That is,
$$ (n+1)!x^n \leq (n+2)! x^{n+1}, $$
or, after dividing through by the left-hand side,
$$ 1 \leq (n+2)x. $$
Thus
$$ n \geq \frac{1}{x} - 2. $$
The smallest integer $n$ satisfying this inequality is
$$ n = \left\lceil \frac{1}{x} - 2 \right\rceil. $$