Numerically summing a divergent series

Question

Say I have a closed form expression for a divergent series and I can calculate as many terms in it as I want. What options have I got to obtain a meaningful result for this divergent series?

Answer 1

Despite having been down-voted this is a perfectly valid question. Divergent (or better: asymptotic) series are a very common phenomenon in physics. People like Michael Berry have spent large parts of their research career on this topic (see e.g. Item 6 on asymptotics, plus links therein, on his homepage). Useful review articles are for instance the ones by G. Dunne and J. Boyd.

The situation suggested in the OP - the series being known in closed form - is not typical, as one will normally know only the first few lowest-order terms. But let's assume we have an asymptotic series associated with a function $f(x)$ for which we write the formal power series

$$ f(x) \sim \sum_{n=1}^\infty a_n x^n \equiv \sum_{n=1}^\infty f_n \; , $$

with $x$ denoting a "small" expansion parameter.

The right hand side is divergent, but we can always truncate the series,

$$ f_N (x) \equiv \sum_{n=1}^N f_n \; ,$$

which clearly is finite. The question then is: How good an approximation (if any) to $f(x)$ are the truncations $f_N(x)$ at a given fixed $x$? One has to consider two cases: (i) The $f_N$ approach $f$ monotonically with increasing $N$ until an extremum closest to $f$ is reached. This will happen at an optimal truncation value, $N_0$. For larger $N$ the series will then "explode". (ii) The same happens with the $f_N$ alternating around $f$, so there is an oscillatory approach followed by the fast deviation. [Technical point: As explained in Dunne's article, the alternating series is Borel summable, while the non-alternating one is not.]

Let me provide a standard, hand-waving argument about the optimal truncation value, $N_0$. In typical examples (perturbation theory, WKB), the divergence of the series is due to factorial growth of the coefficients, $a_n \sim n!$, at least for large $n$. We may assume that the series starts to "explode" when the next term added is of the same magnitude as the previous one, so

$$ 1 = \frac{f_n}{f_{n-1}} = \frac{n! \, x^n}{(n-1)! \, x^{n-1}} = nx \; . $$

This suggests that, as a rule of thumb, the optimal truncation value is the inverse of the expansion parameter,

$$ N_0 = 1/x \; .$$

This is nicely seen in Fig. 1 of Boyd's review. To proceed numerically, just plot the truncations $f_N$ versus $N$ and look for the extremum before things "explode". This will provide your "best" numerical answer.

A final remark: Quantum field theories like QED and QCD heavily rely on perturbation theory in the coupling constants, $\alpha = 1/137$ and $\alpha_S$. The perturbative expansions are not known in closed form, but coefficients tend to grow factorially (with the number of Feynman diagrams). The rule of thumb above suggests that QED perturbation theory breaks down after the 137th order. This is an academic problem, though, as already at low orders the (theoretically unknown) error terms are exceedingly small. It's basically experiment that comes to rescue: the anomalous magnetic moment of the electron, $g-2$, for instance, is known to fifth order in the coupling and coincides with the experimental value to 1 part in a trillion - despite the fact that the QED perturbation series is only asymptotic. In QCD, where $\alpha_S \sim 1/3$ for some processes (depending on resolution/energy) the situation is somewhat more tricky. I'm not an expert, but, as far as I know, people continue to think about the asymptotic character of QCD perturbation theory and its remedies.