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 $$ b_g=\sum_{t=0}^T\left(\sum_{k=0}^\infty \frac{s_{(k,t)}}{g^k}\right)\frac{1}{(\log g)^t} $$ $T$ is a finite integer, the logarithm powers are finite. The coefficients $s_{(k,t)}$ are arbitrary coefficients, in principle not related to one another. As an example, the succession for $T=2$ would be $$ \sum_{k=0}^\infty\frac{s_{(k,0)}}{g^k}+\frac{1}{\log g}\sum_{k=0}^\infty\frac{s_{(k,1)}}{g^k}+\frac{1}{(\log g)^2}\sum_{k=0}^\infty\frac{s_{(k,2)}}{g^k}. $$ Now convergence is way slower: as an example, for $T=1$, terms with no $1/g$ powers attached ($k=0$) 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!
