Convergence acceleration technique for $zeta(4)$ (or $eta(4)$) via creative telescoping?












16












$begingroup$



Question




  1. Is it already known whether the $zeta(4):=sum_{n=1}^{infty}1/n^4$ accelerated convergence series $(1)$, proved for instance in [1, Corollaire 5.3], could be obtained by a similar technique to the ones explained by Alf van der Poorten in [2, section 1] for $zeta(3)$ and $zeta(2)$?
    $$zeta(4)=frac{36}{17}sum_{n=1}^{infty}frac{1}{n^{4}binom{2n}{n}}.tag{1}$$

  2. (a) In other words, does there exist a pair of functions $F(n,k), G(n,k)$ obeying equation
    $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)tag{$ast$}$$
    from which $(1)$ can be proved? That is, is it possible to transform the defining series for $zeta(4):=sum_{n=1}^{infty}1/n^4$ by means of the Wilf-Zeilberger method (or the Markov-WZ Method) into the faster series $(1)$? (b) Most likely there isn't any such a pair $(F, G)$, but I do not have the means to use these methods on my own.




Short description of section 1 of Alf van der Poorten's paper



The defining series for $zeta(3):=sum_{n=1}^{infty}1/n^3$ and $zeta(2):=sum_{n=1}^{infty}1/n^2$ are accelerated resulting in



begin{equation*}
zeta (2)=3sum_{n=1}^{infty }
frac{1}{n^{2}binom{2n}{n}},tag{2}
end{equation*}



begin{equation*}
zeta (3)=frac{5}{2}sum_{n=1}^{infty }
frac{(-1)^{n-1}}{n^{3}binom{2n}{n}}tag{3}.
end{equation*}



For instance, $(3)$ follows from the identity



begin{equation*}
sum_{n=1}^{N}frac{1}{n^{3}}-2sum_{n=1}^{N}frac{left( -1right) ^{n-1}}{n^{3}binom{2n}{n}}=sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{N+k}{k}binom{N}{k}}-sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{2k}{k}}tag{4},
end{equation*}



by letting $Nrightarrow infty $ and noticing that



begin{equation*}
lim_{Ntoinfty}sum_{k=1}^{infty}frac{(-1)^{k}}{2k^{3}binom{N+k}{k}binom{N}{k}}=0.
end{equation*}



Equality $(4)$ can be explained as follows:




  1. Write
    begin{equation*}
    X_{n,k}=frac{(-1)^{k-1}}{k^{2}binom{n+k}{k}binom{n-1}{k}},qquad D_{n,k}=frac{(-1)^{k}}{n^{2}binom{n+k}{k}binom{n-1}{k}}qquad k<n.
    end{equation*}


  2. Notice that $$X_{n,k}=D_{n,k-1}-D_{n,k}.tag{5}$$ Hence
    begin{eqnarray*}
    sum_{k=1}^{n-1}frac{X_{n,k}}{n} &=&sum_{k=1}^{n-1}left( frac{D_{n,k-1}}{
    n}-frac{D_{n,k}}{n}right) =frac{D_{n,0}}{n}-frac{D_{n,n-1}}{n} \
    &=&frac{1}{n^{3}}-2frac{left( -1right) ^{n-1}}{n^{3}binom{2n}{n}},qquadfrac{D_{n,0}}{n} =frac{1}{n^{3}},quad frac{D_{n,n-1}}{n}=2frac{
    left( -1right) ^{n-1}}{n^{3}binom{2n}{n}}
    end{eqnarray*}


  3. Sum over $k$, $1leq
    kleq n-1$

    begin{equation*}
    sum_{k=1}^{n-1}X_{n,k}=sum_{k=1}^{n-1}left( D_{n,k-1}-D_{n,k}right)
    =D_{n,0}-D_{n,n-1}.
    end{equation*}


  4. Now, summing over $n$, $1leq nleq N$, and noticing that
    begin{equation*}
    frac{X_{n,k}}{n}=E_{n,k}-E_{n-1,k},qquad E_{n,k}=frac{(-1)^{k}}{2k^{3}binom{n+k}{k}binom{n}{k}},tag{6}
    end{equation*}

    we obtain
    begin{equation*}
    sum_{k=1}^{N-1}sum_{n=k+1}^{N}frac{X_{n,k}}{n}=sum_{k=1}^{N-1}
    sum_{n=k+1}^{N}left( E_{n,k}-E_{n-1,k}right) =sum_{k=1}^{N}left(
    E_{N,k}-E_{k,k}right).
    end{equation*}


  5. So, on the one hand
    begin{eqnarray*}
    sum_{n=1}^{N}sum_{k=1}^{n-1}frac{X_{n,k}}{n} &=&sum_{n=1}^{N}frac{1}{
    n^{3}}-2sum_{n=1}^{N}frac{left( -1right) ^{n-1}}{n^{3}binom{2n}{n}},tag{7}
    end{eqnarray*}

    and on the other hand
    begin{eqnarray*}
    sum_{n=1}^{N}sum_{k=1}^{n-1}frac{X_{n,k}}{n}
    &=&sum_{k=1}^{N}E_{N,k}-sum_{k=1}^{N}E_{k,k} \
    &=&sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{N+k}{k}binom{N}{k}}
    -sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{2k}{k}}.tag{8}
    end{eqnarray*}

    The identity $(4)$ follows.


Remarks




  1. The combination of equations $(5)$ and $(6)$ forms an identity of the form $(ast)$, which is equation $(6.1.2)$ of [3, chapter 6] (Zeilberger's Algorithm).

  2. As for $(2)$, [2, section 1] actually explains how to accelerate $eta(2):=sum_{n=1}^{infty }(-1)^{n-1}/n^{2}$ and obtain $(2)$, using the
    relation $eta(s) = left(1-2^{1-s}right) zeta(s)$. As such, if feasible, I expect that accelerating $eta(4)$ might be easier than $zeta(4)$.


References




  1. Henri Cohen, Généralisation d'une Construction de R. Apéry

  2. Alfred van der Poorten, Some wonderful formulae... Footnotes to Apery's proof of the irrationality of $zeta(3)$

  3. Marko Petkovsek, Herbert Wilf, Doron Zeilberger, A = B










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  • $begingroup$
    what is the question?
    $endgroup$
    – Masacroso
    May 14 '17 at 19:42










  • $begingroup$
    @Masacroso In short, can (1) be derived from an identity similar to (4)?
    $endgroup$
    – Américo Tavares
    May 14 '17 at 19:45






  • 1




    $begingroup$
    See also this fast converging series en.wikipedia.org/wiki/…
    $endgroup$
    – reuns
    May 19 '17 at 20:53


















16












$begingroup$



Question




  1. Is it already known whether the $zeta(4):=sum_{n=1}^{infty}1/n^4$ accelerated convergence series $(1)$, proved for instance in [1, Corollaire 5.3], could be obtained by a similar technique to the ones explained by Alf van der Poorten in [2, section 1] for $zeta(3)$ and $zeta(2)$?
    $$zeta(4)=frac{36}{17}sum_{n=1}^{infty}frac{1}{n^{4}binom{2n}{n}}.tag{1}$$

  2. (a) In other words, does there exist a pair of functions $F(n,k), G(n,k)$ obeying equation
    $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)tag{$ast$}$$
    from which $(1)$ can be proved? That is, is it possible to transform the defining series for $zeta(4):=sum_{n=1}^{infty}1/n^4$ by means of the Wilf-Zeilberger method (or the Markov-WZ Method) into the faster series $(1)$? (b) Most likely there isn't any such a pair $(F, G)$, but I do not have the means to use these methods on my own.




Short description of section 1 of Alf van der Poorten's paper



The defining series for $zeta(3):=sum_{n=1}^{infty}1/n^3$ and $zeta(2):=sum_{n=1}^{infty}1/n^2$ are accelerated resulting in



begin{equation*}
zeta (2)=3sum_{n=1}^{infty }
frac{1}{n^{2}binom{2n}{n}},tag{2}
end{equation*}



begin{equation*}
zeta (3)=frac{5}{2}sum_{n=1}^{infty }
frac{(-1)^{n-1}}{n^{3}binom{2n}{n}}tag{3}.
end{equation*}



For instance, $(3)$ follows from the identity



begin{equation*}
sum_{n=1}^{N}frac{1}{n^{3}}-2sum_{n=1}^{N}frac{left( -1right) ^{n-1}}{n^{3}binom{2n}{n}}=sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{N+k}{k}binom{N}{k}}-sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{2k}{k}}tag{4},
end{equation*}



by letting $Nrightarrow infty $ and noticing that



begin{equation*}
lim_{Ntoinfty}sum_{k=1}^{infty}frac{(-1)^{k}}{2k^{3}binom{N+k}{k}binom{N}{k}}=0.
end{equation*}



Equality $(4)$ can be explained as follows:




  1. Write
    begin{equation*}
    X_{n,k}=frac{(-1)^{k-1}}{k^{2}binom{n+k}{k}binom{n-1}{k}},qquad D_{n,k}=frac{(-1)^{k}}{n^{2}binom{n+k}{k}binom{n-1}{k}}qquad k<n.
    end{equation*}


  2. Notice that $$X_{n,k}=D_{n,k-1}-D_{n,k}.tag{5}$$ Hence
    begin{eqnarray*}
    sum_{k=1}^{n-1}frac{X_{n,k}}{n} &=&sum_{k=1}^{n-1}left( frac{D_{n,k-1}}{
    n}-frac{D_{n,k}}{n}right) =frac{D_{n,0}}{n}-frac{D_{n,n-1}}{n} \
    &=&frac{1}{n^{3}}-2frac{left( -1right) ^{n-1}}{n^{3}binom{2n}{n}},qquadfrac{D_{n,0}}{n} =frac{1}{n^{3}},quad frac{D_{n,n-1}}{n}=2frac{
    left( -1right) ^{n-1}}{n^{3}binom{2n}{n}}
    end{eqnarray*}


  3. Sum over $k$, $1leq
    kleq n-1$

    begin{equation*}
    sum_{k=1}^{n-1}X_{n,k}=sum_{k=1}^{n-1}left( D_{n,k-1}-D_{n,k}right)
    =D_{n,0}-D_{n,n-1}.
    end{equation*}


  4. Now, summing over $n$, $1leq nleq N$, and noticing that
    begin{equation*}
    frac{X_{n,k}}{n}=E_{n,k}-E_{n-1,k},qquad E_{n,k}=frac{(-1)^{k}}{2k^{3}binom{n+k}{k}binom{n}{k}},tag{6}
    end{equation*}

    we obtain
    begin{equation*}
    sum_{k=1}^{N-1}sum_{n=k+1}^{N}frac{X_{n,k}}{n}=sum_{k=1}^{N-1}
    sum_{n=k+1}^{N}left( E_{n,k}-E_{n-1,k}right) =sum_{k=1}^{N}left(
    E_{N,k}-E_{k,k}right).
    end{equation*}


  5. So, on the one hand
    begin{eqnarray*}
    sum_{n=1}^{N}sum_{k=1}^{n-1}frac{X_{n,k}}{n} &=&sum_{n=1}^{N}frac{1}{
    n^{3}}-2sum_{n=1}^{N}frac{left( -1right) ^{n-1}}{n^{3}binom{2n}{n}},tag{7}
    end{eqnarray*}

    and on the other hand
    begin{eqnarray*}
    sum_{n=1}^{N}sum_{k=1}^{n-1}frac{X_{n,k}}{n}
    &=&sum_{k=1}^{N}E_{N,k}-sum_{k=1}^{N}E_{k,k} \
    &=&sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{N+k}{k}binom{N}{k}}
    -sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{2k}{k}}.tag{8}
    end{eqnarray*}

    The identity $(4)$ follows.


Remarks




  1. The combination of equations $(5)$ and $(6)$ forms an identity of the form $(ast)$, which is equation $(6.1.2)$ of [3, chapter 6] (Zeilberger's Algorithm).

  2. As for $(2)$, [2, section 1] actually explains how to accelerate $eta(2):=sum_{n=1}^{infty }(-1)^{n-1}/n^{2}$ and obtain $(2)$, using the
    relation $eta(s) = left(1-2^{1-s}right) zeta(s)$. As such, if feasible, I expect that accelerating $eta(4)$ might be easier than $zeta(4)$.


References




  1. Henri Cohen, Généralisation d'une Construction de R. Apéry

  2. Alfred van der Poorten, Some wonderful formulae... Footnotes to Apery's proof of the irrationality of $zeta(3)$

  3. Marko Petkovsek, Herbert Wilf, Doron Zeilberger, A = B










share|cite|improve this question











$endgroup$












  • $begingroup$
    what is the question?
    $endgroup$
    – Masacroso
    May 14 '17 at 19:42










  • $begingroup$
    @Masacroso In short, can (1) be derived from an identity similar to (4)?
    $endgroup$
    – Américo Tavares
    May 14 '17 at 19:45






  • 1




    $begingroup$
    See also this fast converging series en.wikipedia.org/wiki/…
    $endgroup$
    – reuns
    May 19 '17 at 20:53
















16












16








16


1



$begingroup$



Question




  1. Is it already known whether the $zeta(4):=sum_{n=1}^{infty}1/n^4$ accelerated convergence series $(1)$, proved for instance in [1, Corollaire 5.3], could be obtained by a similar technique to the ones explained by Alf van der Poorten in [2, section 1] for $zeta(3)$ and $zeta(2)$?
    $$zeta(4)=frac{36}{17}sum_{n=1}^{infty}frac{1}{n^{4}binom{2n}{n}}.tag{1}$$

  2. (a) In other words, does there exist a pair of functions $F(n,k), G(n,k)$ obeying equation
    $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)tag{$ast$}$$
    from which $(1)$ can be proved? That is, is it possible to transform the defining series for $zeta(4):=sum_{n=1}^{infty}1/n^4$ by means of the Wilf-Zeilberger method (or the Markov-WZ Method) into the faster series $(1)$? (b) Most likely there isn't any such a pair $(F, G)$, but I do not have the means to use these methods on my own.




Short description of section 1 of Alf van der Poorten's paper



The defining series for $zeta(3):=sum_{n=1}^{infty}1/n^3$ and $zeta(2):=sum_{n=1}^{infty}1/n^2$ are accelerated resulting in



begin{equation*}
zeta (2)=3sum_{n=1}^{infty }
frac{1}{n^{2}binom{2n}{n}},tag{2}
end{equation*}



begin{equation*}
zeta (3)=frac{5}{2}sum_{n=1}^{infty }
frac{(-1)^{n-1}}{n^{3}binom{2n}{n}}tag{3}.
end{equation*}



For instance, $(3)$ follows from the identity



begin{equation*}
sum_{n=1}^{N}frac{1}{n^{3}}-2sum_{n=1}^{N}frac{left( -1right) ^{n-1}}{n^{3}binom{2n}{n}}=sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{N+k}{k}binom{N}{k}}-sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{2k}{k}}tag{4},
end{equation*}



by letting $Nrightarrow infty $ and noticing that



begin{equation*}
lim_{Ntoinfty}sum_{k=1}^{infty}frac{(-1)^{k}}{2k^{3}binom{N+k}{k}binom{N}{k}}=0.
end{equation*}



Equality $(4)$ can be explained as follows:




  1. Write
    begin{equation*}
    X_{n,k}=frac{(-1)^{k-1}}{k^{2}binom{n+k}{k}binom{n-1}{k}},qquad D_{n,k}=frac{(-1)^{k}}{n^{2}binom{n+k}{k}binom{n-1}{k}}qquad k<n.
    end{equation*}


  2. Notice that $$X_{n,k}=D_{n,k-1}-D_{n,k}.tag{5}$$ Hence
    begin{eqnarray*}
    sum_{k=1}^{n-1}frac{X_{n,k}}{n} &=&sum_{k=1}^{n-1}left( frac{D_{n,k-1}}{
    n}-frac{D_{n,k}}{n}right) =frac{D_{n,0}}{n}-frac{D_{n,n-1}}{n} \
    &=&frac{1}{n^{3}}-2frac{left( -1right) ^{n-1}}{n^{3}binom{2n}{n}},qquadfrac{D_{n,0}}{n} =frac{1}{n^{3}},quad frac{D_{n,n-1}}{n}=2frac{
    left( -1right) ^{n-1}}{n^{3}binom{2n}{n}}
    end{eqnarray*}


  3. Sum over $k$, $1leq
    kleq n-1$

    begin{equation*}
    sum_{k=1}^{n-1}X_{n,k}=sum_{k=1}^{n-1}left( D_{n,k-1}-D_{n,k}right)
    =D_{n,0}-D_{n,n-1}.
    end{equation*}


  4. Now, summing over $n$, $1leq nleq N$, and noticing that
    begin{equation*}
    frac{X_{n,k}}{n}=E_{n,k}-E_{n-1,k},qquad E_{n,k}=frac{(-1)^{k}}{2k^{3}binom{n+k}{k}binom{n}{k}},tag{6}
    end{equation*}

    we obtain
    begin{equation*}
    sum_{k=1}^{N-1}sum_{n=k+1}^{N}frac{X_{n,k}}{n}=sum_{k=1}^{N-1}
    sum_{n=k+1}^{N}left( E_{n,k}-E_{n-1,k}right) =sum_{k=1}^{N}left(
    E_{N,k}-E_{k,k}right).
    end{equation*}


  5. So, on the one hand
    begin{eqnarray*}
    sum_{n=1}^{N}sum_{k=1}^{n-1}frac{X_{n,k}}{n} &=&sum_{n=1}^{N}frac{1}{
    n^{3}}-2sum_{n=1}^{N}frac{left( -1right) ^{n-1}}{n^{3}binom{2n}{n}},tag{7}
    end{eqnarray*}

    and on the other hand
    begin{eqnarray*}
    sum_{n=1}^{N}sum_{k=1}^{n-1}frac{X_{n,k}}{n}
    &=&sum_{k=1}^{N}E_{N,k}-sum_{k=1}^{N}E_{k,k} \
    &=&sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{N+k}{k}binom{N}{k}}
    -sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{2k}{k}}.tag{8}
    end{eqnarray*}

    The identity $(4)$ follows.


Remarks




  1. The combination of equations $(5)$ and $(6)$ forms an identity of the form $(ast)$, which is equation $(6.1.2)$ of [3, chapter 6] (Zeilberger's Algorithm).

  2. As for $(2)$, [2, section 1] actually explains how to accelerate $eta(2):=sum_{n=1}^{infty }(-1)^{n-1}/n^{2}$ and obtain $(2)$, using the
    relation $eta(s) = left(1-2^{1-s}right) zeta(s)$. As such, if feasible, I expect that accelerating $eta(4)$ might be easier than $zeta(4)$.


References




  1. Henri Cohen, Généralisation d'une Construction de R. Apéry

  2. Alfred van der Poorten, Some wonderful formulae... Footnotes to Apery's proof of the irrationality of $zeta(3)$

  3. Marko Petkovsek, Herbert Wilf, Doron Zeilberger, A = B










share|cite|improve this question











$endgroup$





Question




  1. Is it already known whether the $zeta(4):=sum_{n=1}^{infty}1/n^4$ accelerated convergence series $(1)$, proved for instance in [1, Corollaire 5.3], could be obtained by a similar technique to the ones explained by Alf van der Poorten in [2, section 1] for $zeta(3)$ and $zeta(2)$?
    $$zeta(4)=frac{36}{17}sum_{n=1}^{infty}frac{1}{n^{4}binom{2n}{n}}.tag{1}$$

  2. (a) In other words, does there exist a pair of functions $F(n,k), G(n,k)$ obeying equation
    $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)tag{$ast$}$$
    from which $(1)$ can be proved? That is, is it possible to transform the defining series for $zeta(4):=sum_{n=1}^{infty}1/n^4$ by means of the Wilf-Zeilberger method (or the Markov-WZ Method) into the faster series $(1)$? (b) Most likely there isn't any such a pair $(F, G)$, but I do not have the means to use these methods on my own.




Short description of section 1 of Alf van der Poorten's paper



The defining series for $zeta(3):=sum_{n=1}^{infty}1/n^3$ and $zeta(2):=sum_{n=1}^{infty}1/n^2$ are accelerated resulting in



begin{equation*}
zeta (2)=3sum_{n=1}^{infty }
frac{1}{n^{2}binom{2n}{n}},tag{2}
end{equation*}



begin{equation*}
zeta (3)=frac{5}{2}sum_{n=1}^{infty }
frac{(-1)^{n-1}}{n^{3}binom{2n}{n}}tag{3}.
end{equation*}



For instance, $(3)$ follows from the identity



begin{equation*}
sum_{n=1}^{N}frac{1}{n^{3}}-2sum_{n=1}^{N}frac{left( -1right) ^{n-1}}{n^{3}binom{2n}{n}}=sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{N+k}{k}binom{N}{k}}-sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{2k}{k}}tag{4},
end{equation*}



by letting $Nrightarrow infty $ and noticing that



begin{equation*}
lim_{Ntoinfty}sum_{k=1}^{infty}frac{(-1)^{k}}{2k^{3}binom{N+k}{k}binom{N}{k}}=0.
end{equation*}



Equality $(4)$ can be explained as follows:




  1. Write
    begin{equation*}
    X_{n,k}=frac{(-1)^{k-1}}{k^{2}binom{n+k}{k}binom{n-1}{k}},qquad D_{n,k}=frac{(-1)^{k}}{n^{2}binom{n+k}{k}binom{n-1}{k}}qquad k<n.
    end{equation*}


  2. Notice that $$X_{n,k}=D_{n,k-1}-D_{n,k}.tag{5}$$ Hence
    begin{eqnarray*}
    sum_{k=1}^{n-1}frac{X_{n,k}}{n} &=&sum_{k=1}^{n-1}left( frac{D_{n,k-1}}{
    n}-frac{D_{n,k}}{n}right) =frac{D_{n,0}}{n}-frac{D_{n,n-1}}{n} \
    &=&frac{1}{n^{3}}-2frac{left( -1right) ^{n-1}}{n^{3}binom{2n}{n}},qquadfrac{D_{n,0}}{n} =frac{1}{n^{3}},quad frac{D_{n,n-1}}{n}=2frac{
    left( -1right) ^{n-1}}{n^{3}binom{2n}{n}}
    end{eqnarray*}


  3. Sum over $k$, $1leq
    kleq n-1$

    begin{equation*}
    sum_{k=1}^{n-1}X_{n,k}=sum_{k=1}^{n-1}left( D_{n,k-1}-D_{n,k}right)
    =D_{n,0}-D_{n,n-1}.
    end{equation*}


  4. Now, summing over $n$, $1leq nleq N$, and noticing that
    begin{equation*}
    frac{X_{n,k}}{n}=E_{n,k}-E_{n-1,k},qquad E_{n,k}=frac{(-1)^{k}}{2k^{3}binom{n+k}{k}binom{n}{k}},tag{6}
    end{equation*}

    we obtain
    begin{equation*}
    sum_{k=1}^{N-1}sum_{n=k+1}^{N}frac{X_{n,k}}{n}=sum_{k=1}^{N-1}
    sum_{n=k+1}^{N}left( E_{n,k}-E_{n-1,k}right) =sum_{k=1}^{N}left(
    E_{N,k}-E_{k,k}right).
    end{equation*}


  5. So, on the one hand
    begin{eqnarray*}
    sum_{n=1}^{N}sum_{k=1}^{n-1}frac{X_{n,k}}{n} &=&sum_{n=1}^{N}frac{1}{
    n^{3}}-2sum_{n=1}^{N}frac{left( -1right) ^{n-1}}{n^{3}binom{2n}{n}},tag{7}
    end{eqnarray*}

    and on the other hand
    begin{eqnarray*}
    sum_{n=1}^{N}sum_{k=1}^{n-1}frac{X_{n,k}}{n}
    &=&sum_{k=1}^{N}E_{N,k}-sum_{k=1}^{N}E_{k,k} \
    &=&sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{N+k}{k}binom{N}{k}}
    -sum_{k=1}^{N}frac{(-1)^{k}}{2k^{3}binom{2k}{k}}.tag{8}
    end{eqnarray*}

    The identity $(4)$ follows.


Remarks




  1. The combination of equations $(5)$ and $(6)$ forms an identity of the form $(ast)$, which is equation $(6.1.2)$ of [3, chapter 6] (Zeilberger's Algorithm).

  2. As for $(2)$, [2, section 1] actually explains how to accelerate $eta(2):=sum_{n=1}^{infty }(-1)^{n-1}/n^{2}$ and obtain $(2)$, using the
    relation $eta(s) = left(1-2^{1-s}right) zeta(s)$. As such, if feasible, I expect that accelerating $eta(4)$ might be easier than $zeta(4)$.


References




  1. Henri Cohen, Généralisation d'une Construction de R. Apéry

  2. Alfred van der Poorten, Some wonderful formulae... Footnotes to Apery's proof of the irrationality of $zeta(3)$

  3. Marko Petkovsek, Herbert Wilf, Doron Zeilberger, A = B







sequences-and-series number-theory summation experimental-mathematics convergence-acceleration






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edited Dec 18 '18 at 21:30







Américo Tavares

















asked May 14 '17 at 19:36









Américo TavaresAmérico Tavares

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  • $begingroup$
    what is the question?
    $endgroup$
    – Masacroso
    May 14 '17 at 19:42










  • $begingroup$
    @Masacroso In short, can (1) be derived from an identity similar to (4)?
    $endgroup$
    – Américo Tavares
    May 14 '17 at 19:45






  • 1




    $begingroup$
    See also this fast converging series en.wikipedia.org/wiki/…
    $endgroup$
    – reuns
    May 19 '17 at 20:53




















  • $begingroup$
    what is the question?
    $endgroup$
    – Masacroso
    May 14 '17 at 19:42










  • $begingroup$
    @Masacroso In short, can (1) be derived from an identity similar to (4)?
    $endgroup$
    – Américo Tavares
    May 14 '17 at 19:45






  • 1




    $begingroup$
    See also this fast converging series en.wikipedia.org/wiki/…
    $endgroup$
    – reuns
    May 19 '17 at 20:53


















$begingroup$
what is the question?
$endgroup$
– Masacroso
May 14 '17 at 19:42




$begingroup$
what is the question?
$endgroup$
– Masacroso
May 14 '17 at 19:42












$begingroup$
@Masacroso In short, can (1) be derived from an identity similar to (4)?
$endgroup$
– Américo Tavares
May 14 '17 at 19:45




$begingroup$
@Masacroso In short, can (1) be derived from an identity similar to (4)?
$endgroup$
– Américo Tavares
May 14 '17 at 19:45




1




1




$begingroup$
See also this fast converging series en.wikipedia.org/wiki/…
$endgroup$
– reuns
May 19 '17 at 20:53






$begingroup$
See also this fast converging series en.wikipedia.org/wiki/…
$endgroup$
– reuns
May 19 '17 at 20:53












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