1) Because the asymptotic "convergence" is relative to a POINT. This point is often g = 0 in perturbation theory. And around that point (g = 0) you can bound the difference between the function and the truncated expansion at given order by the the first omitted term.

Asymptotic series (around a point): Convergence around that SINGLE point for fixed order (number of terms).

Convergent series: Convergence in the limit of infinite series for ALL fixed points.

2) A QFT amplitude has the form

\[A\sim\sum_{n=0}^{\infty} a_n=\sum_{n=0}^{\infty} (g/2\pi)^n\,(2n-1)!!\]

Because the factorial grows very rapidly, the series does not converge (absolutely). However, it does converge asymptotically around g = 0. The condition is:

\[\lim_{g\to 0} \frac{A-\sum_{n=0}^{N}a_n}{a_N} = 0\]

Because of the limit (which should actually be $g\to 0^+$), we can replace the numerator with the dominant term, so that

\[\lim_{g\to 0} \frac{A-\sum_{n=0}^{N}a_n}{a_N} = \frac{a_{N+1}}{a_N}=(2N+1)\,(g/2\pi) =0\]

Now, asymptotic convergence around g = 0 implies that for a finite \(g = g_0\) (different from 0), beyond certain truncation N, the relative error gets worse. This N is given by the condition

\[\frac{A-\sum_{n=0}^{N}a_n}{a_N} < 1\]

which translates in \(N \sim \pi/g\) . In QED (g=1/137), this is \(N\sim 430\), whereas the experimental accuracy is restricted to N = 4 .

Edit: I misunderstood the question: if you take the first N terms, the estimate of your error is given by the N+1 term, as long as \(N < \pi/g\)

**Toy integral analogy:**

Consider the integral:

\[I(g)=\int _0^{\infty}\frac{e^{-t}}{1+g\,t}\, dt\]

where g is non-negative.

For \(g\to 0^+\) we can replace the denominator by its power series expansion and integrate term by term, getting:

\[\lim _{g\to 0^+}\, I(g) = \sum _{n=0}^{\infty}\, (-1)^nn!\, g^n\]

The series does not converge because of the factorial, but due to the previous equality, the series \[\sum _{n=0}^{N}\, (-1)^nn!\, g^n\] is asymptotic to I(g) for \(g\to 0^+\) for all N.

The reason for the lack of convergence of the series \[\sum _{n=0}^{\infty}\, (-1)^nn!\, g^n\] is that the expansion in power series of \(\frac{1}{1+g\,t}\) is only convergent for t lower than 1/g (**honest question:** **what's the analogous in QFT??**). Another way to see this, it's that for negative g, the integrand has a non-integrable singularity at t equal to -1/g, so that the integral isn't defined. This is essentially Dyson's argument for the nonconvergence of series in QFT (from Wikipedia): if the series were convergent in a disk of non-null radius centered at g=0, it should obviously include negative values for the coupling constant g. But if the coupling constant were negative, this would be equivalent to the Coulomb force constant being negative. This would "reverse" the electromagnetic interaction so that like charges would attract and unlike charges would repel. This would render the vacuum unstable against decay into a cluster of electrons on one side of the universe and a cluster of positrons on the other side of the universe. Because the theory is 'sick' for any negative value of the coupling constant, the theory doesn;t exist and the series cannot converge, but is an asymptotic series.

3) I cannot improve this link: https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/