I have a numerical question regarding acceleration of a succession.

A preliminary: suppose that I have a succession $ a_g$ that, for high $ g$ , asymptotically goes as $ $ a_g=s_0+\frac{s_1}g+\frac{s_2}{g^2}+…=\sum_{k=0}^\infty \frac{s_k}{g^k}. $ $ I am interested in computing the coefficients $ s_k$ , but in particular I am interested in computing the leading coefficient $ s_0$ (that would also be the limit of the succession as $ g\to\infty$ ). A way to accelerate this convergence is given by the Richardson transform: by defining the succession $ $ a_g^{(N)}=\sum_{n=0}^N(-1)^{n+N}\frac{(g+n)^N}{n!(N-n)!}a_{g+n}, $ $ it can be proven that $ a_g^{(N)}$ goes to the same limit as $ a_g$ , but the succession is accelerated as the asymptotic behavior is $ $ a_g^{(N)}\simeq s_0+\sum_{k=N+1}^\infty \frac{d_k}{g^k}. $ $ This works nicely and gives very good numerical results in the examples I’ve used.

The problem is that I’m now working with more general successions, of the form $ $ b_g=\sum_{t=0}^T\left(\sum_{k=0}^\infty s_{(k,t)}\right)\frac{1}{(\log g)^t} $ $ $ T$ is a finite integer, the logarithm powers are finite. Now succession is way slower: as an example, for $ T=1$ , terms with no $ 1/g$ powers attached are $ $ s_{(0,0)}+\frac{s_{(0,1)}}{\log g}. $ $ If I want to compute $ s_{(0,0)}$ by computing $ b_g$ for high $ g$ , I get very slow convergence, as $ (\log g)^{-1}$ goes to zero very slowly. The standard Richardson transform does not really work here.

The question: is there a generalization of this Richardson transform to successions like the $ b_g$ succession?

Thanks everybody!