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I try to compare several method of sum acceleration series and testing in the following Double sum $$\sum _{k=0}^{\infty } \sum _{j=0}^{\infty } (-1)^{j+k} \log ^2(j+k+1)=-0.00652533458496$$ (Ithink this is the true Value but here begin the troubles Mathematica gives the values using WynnEpsilon complexInfinity and if I Use levinUtrasform I get 5.1729515919603532132234865492317838383764158033761822568 and Maple gives nothing , if I change the values I get (using my method) $$\sum _{k=0}^{\infty } \sum _{j=0}^{\infty } \left(-\frac{1}{6}\right)^{j+k} \log ^2(j+k+1)=-0.087178770080932$$ Maples gives 1.05795918367346939 , Mathematica gives 0.020656409926403872149 and the levinUtrasform -0.087178770080932183760585045710101 could you show what values are correct ??

user547221
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  • What is the LevinUTransform ? – Jean Marie May 20 '21 at 20:48
  • How does the double series converge when for each $k$, the series $\sum_{j=0}^\infty (-1)^{j+k} \log^2(j+k+1)$ diverges? – Mark Viola May 20 '21 at 20:50
  • This question, like your past questions, are non motivated (why are you interested in this sum ? What is you knowledge on it ; you probably already have track ?). Give us context, what you know, what you don't want, etc. My recent experience with one of the last questions you have asked was an extreme difficulty to communicate exactly what you wanted. – Jean Marie May 20 '21 at 20:59
  • the Levin u-transformation is mentioned as a good universal technique for convergence acceleration. .the series$$\sum _{k=0}^{\infty } \sum _{j=0}^{\infty } (-1)^{j+k} \log (j+1) \log (k+1)=0.050981734929135662$$ converge – user547221 May 20 '21 at 21:43
  • https://math.stackexchange.com/questions/6625/levins-u-transformation – user547221 May 20 '21 at 21:44
  • Thanks for your answers. I wasn't aware of them because you hadn't prefixed them by @ followed by my pseudo. – Jean Marie May 21 '21 at 06:08

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