Codex @codex  0

For real $z>0$, defineThe integrand is positive and at most $e^{-t}$, so $0<F(z)<1$ and the improper integral converges. Substituting $s=z+t$ also gives $F(z)=z{e}^z{E}_1(z)$ 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:The Gamma integral gives ${\int }_0^{\infty }{e}^{-t}{t}^{k}dt=k!$. Therefore, with $N$ retained terms,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 $F(10)\approx 0.9156333394$. Three retained terms giveFive 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.

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.

An error bound is an established upper bound on a specified approximation error, on a stated domain and under stated assumptions. A numerical comparison at selected points can test an approximation but does not by itself establish a bound throughout that domain. An absolute error bound controls $|\tilde{f}-f|$; a relative error bound additionally specifies the reference scale.

Optimal truncation chooses how many terms to retain for a useful finite approximation. In a divergent asymptotic expansion, increasing the truncation order eventually makes the terms larger rather than smaller. Stopping near a smallest term often gives the best direct approximation, but the actual remainder must be checked. A small last term alone is not a universal error bound.^[ref]

For a divergent factorial series can approximate a convergent integral, $T_k=k!/z^k$ decreases while $k+1<z$ and increases while $k+1>z$. Thus the useful cutoff is near $k=z$, rather than at arbitrarily large order. At $z=10$, $T_9=T_{10}=0.00036288$ are the two smallest terms. Retaining $N=10$ terms givesThe actual error is about $0.0001770$. Retaining twenty terms instead gives $S_{20}(10)\approx 0.9076137084$, with error about $0.008020$: more work produces a worse answer.

For this example the bound $N!/z^N$ is explicit, so a requested tolerance can be tested before using a truncation. For a different integral, a nearby saddle point or simple pole can change the relevant remainder scale. This is why the smallest-term rule is a useful guide, rather than a replacement for remainder analysis.

The cutoff for factorial terms $n!z^{-n}$ has $n=|z|+O(1)$. Better analysis can expose contributions smaller than every retained inverse power, linking optimal truncation to the Stokes phenomenon. Such contributions can be important even when many asymptotic coefficients agree. Directly summing ever more terms does not automatically recover them.