= A divergent factorial series can approximate a convergent integral
{title2=$F(z)=\int_0^\infty e^{-t}/(1+t/z)\,dt$}
For real $z>0$, define
$$
F(z)=\int_0^\infty\frac{e^{-t}}{1+t/z}\,dt.
$$
The <integrand> is positive and at most $e^{-t}$, so $0<F(z)<1$ and the <improper integral> converges. Substituting $s=z+t$ also gives $F(z)=ze^zE_1(z)$ in terms of the <exponential integral>; the original representation makes positivity and the large-$z$ behavior transparent.
Apply the exact finite <geometric series> identity inside the <integral>, without assuming that an infinite <series> can be integrated term by term:
$$
\frac1{1+t/z}
=\sum_{k=0}^{N-1}(-1)^k\frac{t^k}{z^k}
+(-1)^N\frac{t^N/z^N}{1+t/z}.
$$
The <Gamma integral> gives $\int_0^\infty e^{-t}t^k\,dt=k!$. Therefore, with $N$ retained terms,
$$
F(z)=S_N(z)+R_N(z),\qquad
S_N(z)=\sum_{k=0}^{N-1}\frac{(-1)^kk!}{z^k},
$$
$$
R_N(z)=\frac{(-1)^N}{z^N}
\int_0^\infty\frac{e^{-t}t^N}{1+t/z}\,dt,
\qquad
\boxed{|R_N(z)|\leq\frac{N!}{z^N}}.
$$
This remainder formula proves the <asymptotic expansion> at every fixed order: $R_N=O(z^{-N})=o(z^{-N+1})$ as $z\to+\infty$. Its sign also gives alternating bounds: an odd $N$ overestimates $F(z)$, and an even $N$ underestimates it.
At $z=10$, an independent evaluation of the <integral> gives $F(10)\approx0.9156333394$. Three retained terms give
$$
S_3(10)=1-0.1+0.02=0.92,
\qquad |F(10)-S_3(10)|<0.006.
$$
Five retained terms give $S_5(10)=0.9164$ with error bounded by $5!/10^5=0.0012$. The actual errors are about $0.004367$ and $0.000767$, respectively. The <error bound> is conservative, but established without relying on the numerical integral evaluation.
\Image[/asymptotic-factorial-fixed-order-accuracy.png]
{title=Fixed-order <asymptotic expansions> improve as the parameter grows}
{description=Original calculations for $F(z)=\int_0^\infty e^{-t}/(1+t/z)\,dt$. Left: the integral and its one-, three- and five-term <partial sums>. Right: their absolute errors, with dotted bounds $N!/z^N$. The independently justified bounds apply for every displayed positive $z$; the <asymptotic expansion> concerns the limit $z\to+\infty$.}
Nevertheless, the formal infinite <series> diverges at every fixed $z>0$. The magnitudes $T_k=k!/z^k$ satisfy $T_{k+1}/T_k=(k+1)/z$, so they eventually increase without bound. In particular, the terms fail the necessary zero-term condition for a convergent <series>. The finite remainder identity remains valid after convergence has failed; this is precisely what makes the example useful.
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