# Convergence acceleration of successions with logarithms

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!