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After I've read Kummer's recipe to get the acceleration of a series, I want to do an example related with the Möbius function $\mu(n)$.

Question. I present my calculations, please A) I would like to know if there are mistakes in how I've used Kummer's acceleration method, specially I need to show that (being required by the method) $$s=\sum_{k=0}^\infty\frac{1}{(k+1)^2}\sum_{j=1}^{k+1}\frac{\mu(j)}{j^3}$$ is convergent; and B) please I would like if there are some method, theoretic or well numeric to get an idea about how was accelerated the series, I say a comparison between the old series $s$, and the new expression $$s=\frac{1}{\zeta(3)}\frac{\pi^2}{6}+\sum_{k=0}^\infty\left( \left( 1-\frac{1/\zeta(3)}{\sum_{j=1}^{k+1}\mu(j)/j^3} \right) \frac{\sum_{j=1}^{k+1}\frac{\mu(j)}{j^3}}{(k+1)^2} \right) .$$ Many thanks.

The calculations that I did to get A) was the specialization of Kummer's method that is explained in this [1], with $a_k=\frac{1}{(k+1)^2}\sum_{j=1}^{k+1}\frac{\mu(j)}{j^3}$, and $c_k=\frac{1}{(k+1)^2}$. From particular values of Dirichlet series one has $$\lim_{k\to\infty}\sum_{j=1}^{k+1}\frac{\mu(j)}{j^3}=\frac{1}{\zeta(3)},$$ and from the method $$s=\frac{1}{\zeta(3)}\frac{\pi^2}{6}+\sum_{k=0}^\infty\left( \left( 1-\frac{1/\zeta(3)}{\sum_{j=1}^{k+1}\mu(j)/j^3} \right) \frac{\sum_{j=1}^{k+1}\frac{\mu(j)}{j^3}}{(k+1)^2} \right). $$ But I need to show previously convergence for the genuine $s$. I try do it using that the convergence absolute of a series implies its convergence. Thus my calculations were using Apostol's Theorem 3.2 to show from $$ \left| \sum_{k=0}^\infty\frac{1}{(k+1)^2}\sum_{j=1}^{k+1}\frac{\mu(j)}{j^3}\right|<\sum_{k=0} ^\infty\frac{1}{(k+1)^2} \left( \zeta(3) -\frac{1}{(k+1)^2}+O \left( \frac{1}{(k+1)^3} \right) \right), $$ that the genuine $s$ is convergent because previous RHS is $\frac{\zeta(3)}{\zeta(2)}-\frac{\zeta(4)}{2}+\sum_{k=0}^\infty\frac{1}{(k+1)^2}O\left( \frac{1}{(k+1)^3} \right)$ and thus $O(1)$. In this last step is where I've doubts about how write a rigurous proof of the convergence of the genuine series $s$. And what I am asking in B) us how to get an idea of the accuracy of the digits from the acceleration that we did, I am saying a comparison of accuracy of the digits of both series, well numeric providing a clai about how more or less is improved the convergence of the seies, or well with a theoretic reasoning.

References:

[1] Convergence Improvement, MathWorld.

[2] Apostol, An Introduction to Analytic Number Theory, Springer (1976).

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    It is trivial to show that $\sum_{k=0}^\infty\frac{1}{(k+1)^2}\sum_{j=1}^{k+1}\frac{\mu(j)}{j^3}=\frac{1}{\zeta(3)}\frac{\pi^2}{6}+\sum_{k=0}^\infty\left( \left( 1-\frac{1/\zeta(3)}{\sum_{j=1}^{k+1}\mu(j)/j^3} \right) \frac{\sum_{j=1}^{k+1}\frac{\mu(j)}{j^3}}{(k+1)^2} \right) $. So what is your question ? – 2017-01-21
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    Thanks much for your attention @user1952009 . I believe that it is trivial because the method tell us that we can accelerate a series by means of linear combinations. My question A) was then if my calculation to deduce the absolute convergence in the hypothesis is right, and B) if it is possible to quantity how is improved/accelerated the convergence of the genuine series. – 2017-01-21
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    No. It is trivial because $\frac{\pi^2}{6} = \zeta(2) = \sum_{k=0}^\infty \frac{1}{(k+1)^2}$ – 2017-01-21
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    Then I'm sorry my thoughts what I said. My way was to read the article from MathWorld and try do a simple example @user1952009 – 2017-01-21

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