Closed form for $sumlimits_{n=1}^{infty}frac{O_{n}^{(p)}}{(2n-1)^{q}}$, with...












0














Consider the sum




$$sum_{n=1}^{infty}frac{O_{n}^{(p)}}{(2n-1)^{q}}text{, with }O_{n}^{(s)}=1+frac{1}{3^{s}}+dots+frac{1}{(2n-1)^{s}}$$




My question is: if there exists some general theorems that allow to represent this sums in terms of the Riemann Zeta function, like Borwein-Borwein-Girgensohn theorem for Euler sums?






sequences-and-series closed-form riemann-zeta euler-sums







share|cite|improve this question
























  • These Euler sums are well-known to be solvable (in terms of $zeta$) for $p+qleq 5$ and in other cases (like $p=q$), see The Bible.
    – Jack D'Aurizio
    2 days ago












  • I'm not sure, which theorem exactly do you have in mind?
    – Isak
    yesterday
















0














Consider the sum




$$sum_{n=1}^{infty}frac{O_{n}^{(p)}}{(2n-1)^{q}}text{, with }O_{n}^{(s)}=1+frac{1}{3^{s}}+dots+frac{1}{(2n-1)^{s}}$$




My question is: if there exists some general theorems that allow to represent this sums in terms of the Riemann Zeta function, like Borwein-Borwein-Girgensohn theorem for Euler sums?






sequences-and-series closed-form riemann-zeta euler-sums







share|cite|improve this question
























  • These Euler sums are well-known to be solvable (in terms of $zeta$) for $p+qleq 5$ and in other cases (like $p=q$), see The Bible.
    – Jack D'Aurizio
    2 days ago












  • I'm not sure, which theorem exactly do you have in mind?
    – Isak
    yesterday














0












0








0


0





Consider the sum




$$sum_{n=1}^{infty}frac{O_{n}^{(p)}}{(2n-1)^{q}}text{, with }O_{n}^{(s)}=1+frac{1}{3^{s}}+dots+frac{1}{(2n-1)^{s}}$$




My question is: if there exists some general theorems that allow to represent this sums in terms of the Riemann Zeta function, like Borwein-Borwein-Girgensohn theorem for Euler sums?






sequences-and-series closed-form riemann-zeta euler-sums







share|cite|improve this question















Consider the sum




$$sum_{n=1}^{infty}frac{O_{n}^{(p)}}{(2n-1)^{q}}text{, with }O_{n}^{(s)}=1+frac{1}{3^{s}}+dots+frac{1}{(2n-1)^{s}}$$




My question is: if there exists some general theorems that allow to represent this sums in terms of the Riemann Zeta function, like Borwein-Borwein-Girgensohn theorem for Euler sums?






sequences-and-series closed-form riemann-zeta euler-sums




sequences-and-series closed-form riemann-zeta euler-sums






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited 2 days ago









mrtaurho

3,99521133




3,99521133










asked 2 days ago









Isak

215




215












  • These Euler sums are well-known to be solvable (in terms of $zeta$) for $p+qleq 5$ and in other cases (like $p=q$), see The Bible.
    – Jack D'Aurizio
    2 days ago












  • I'm not sure, which theorem exactly do you have in mind?
    – Isak
    yesterday


















  • These Euler sums are well-known to be solvable (in terms of $zeta$) for $p+qleq 5$ and in other cases (like $p=q$), see The Bible.
    – Jack D'Aurizio
    2 days ago












  • I'm not sure, which theorem exactly do you have in mind?
    – Isak
    yesterday
















These Euler sums are well-known to be solvable (in terms of $zeta$) for $p+qleq 5$ and in other cases (like $p=q$), see The Bible.
– Jack D'Aurizio
2 days ago






These Euler sums are well-known to be solvable (in terms of $zeta$) for $p+qleq 5$ and in other cases (like $p=q$), see The Bible.
– Jack D'Aurizio
2 days ago














I'm not sure, which theorem exactly do you have in mind?
– Isak
yesterday




I'm not sure, which theorem exactly do you have in mind?
– Isak
yesterday










1 Answer
1






active

oldest

votes


















1














Changing your notation a bit
and playing around.



If
$O_{n}(s)
=1+frac{1}{3^{s}}+dots+frac{1}{(2n-1)^{s}}
=sum_{k=1}^n frac1{(2k-1)^s}
$,
$O(s)
=O_{infty}(s)
$,
and
$s(p, q)
=sum_{n=1}^{infty}dfrac{O_{n}(p)}{(2n-1)^{q}}
$
then



$s(p, q)+s(q, p)
=O(p)O(q)+O(p+q)
$.



Proof.



$begin{array}\
s(p, q)
&=sum_{n=1}^{infty}dfrac{O_{n}(p)}{(2n-1)^{q}}\
&=sum_{n=1}^{infty}dfrac{sum_{k=1}^n frac1{(2k-1)^p}}{(2n-1)^{q}}\
&=sum_{n=1}^{infty}sum_{k=1}^n dfrac1{(2k-1)^p}dfrac{1}{(2n-1)^{q}}\
&=sum_{k=1}^{infty}sum_{n=k}^{infty} dfrac1{(2k-1)^p}dfrac{1}{(2n-1)^{q}}\
&=sum_{k=1}^{infty}dfrac1{(2k-1)^p}sum_{n=k}^{infty} dfrac{1}{(2n-1)^{q}}\
&=sum_{k=1}^{infty}dfrac1{(2k-1)^p}left(sum_{n=1}^{infty} dfrac{1}{(2n-1)^{q}}-sum_{n=1}^{k-1} dfrac{1}{(2n-1)^{q}}right)\
&=sum_{k=1}^{infty}dfrac1{(2k-1)^p}left(O(q)-O_{k-1}(q)right)\
&=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}O_{k-1}(q)\
&=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}(O_{k}(q)-frac1{(2k-1)^q})\
&=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}O_{k}(q)+sum_{k=1}^{infty}dfrac1{(2k-1)^p}dfrac1{(2k-1)^q}\
&=O(p)O(q)-s(q, p)+sum_{k=1}^{infty}dfrac1{(2k-1)^{p+q}}\
&=O(p)O(q)-s(q, p)+O(p+q)
\
end{array}
$



Some additional
definitions and stuff
which might be useful.



$Z_{n}(s)
=sum_{k=1}^n frac1{k^s}
$



$E_{n}(s)
=sum_{k=1}^n frac1{(2k)^s}
=frac1{2^s}sum_{k=1}^n frac1{k^s}
=frac1{2^s}Z_{n}(s)
$



$O_{n}(s)+E_{n}(s)
=sum_{k=1}^{2n} frac1{k^s}
=Z_{2n}(s)
$
so
$O_{n}(s)
=Z_{2n}(s)-E_{n}(s)
=Z_{2n}(s)-frac1{2^s}Z_{n}(s)
$



$O(s)
=O_{infty}(s)
$,
$E(s)
=E_{infty}(s)
$,
$Z(s)
=zeta(s)
=Z_{infty}(s)
$.



$O(s)
=Z(s)-E(s)
=Z(s)-frac1{2^s}Z(s)
=(1-2^{-s})Z(s)
$






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Changing your notation a bit
    and playing around.



    If
    $O_{n}(s)
    =1+frac{1}{3^{s}}+dots+frac{1}{(2n-1)^{s}}
    =sum_{k=1}^n frac1{(2k-1)^s}
    $    ,
    $O(s)
    =O_{infty}(s)
    $    ,
    and
    $s(p, q)
    =sum_{n=1}^{infty}dfrac{O_{n}(p)}{(2n-1)^{q}}
    $
    then



    $s(p, q)+s(q, p)
    =O(p)O(q)+O(p+q)
    $    .



    Proof.



    $begin{array}\
    s(p, q)
    &=sum_{n=1}^{infty}dfrac{O_{n}(p)}{(2n-1)^{q}}\
    &=sum_{n=1}^{infty}dfrac{sum_{k=1}^n frac1{(2k-1)^p}}{(2n-1)^{q}}\
    &=sum_{n=1}^{infty}sum_{k=1}^n dfrac1{(2k-1)^p}dfrac{1}{(2n-1)^{q}}\
    &=sum_{k=1}^{infty}sum_{n=k}^{infty} dfrac1{(2k-1)^p}dfrac{1}{(2n-1)^{q}}\
    &=sum_{k=1}^{infty}dfrac1{(2k-1)^p}sum_{n=k}^{infty} dfrac{1}{(2n-1)^{q}}\
    &=sum_{k=1}^{infty}dfrac1{(2k-1)^p}left(sum_{n=1}^{infty} dfrac{1}{(2n-1)^{q}}-sum_{n=1}^{k-1} dfrac{1}{(2n-1)^{q}}right)\
    &=sum_{k=1}^{infty}dfrac1{(2k-1)^p}left(O(q)-O_{k-1}(q)right)\
    &=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}O_{k-1}(q)\
    &=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}(O_{k}(q)-frac1{(2k-1)^q})\
    &=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}O_{k}(q)+sum_{k=1}^{infty}dfrac1{(2k-1)^p}dfrac1{(2k-1)^q}\
    &=O(p)O(q)-s(q, p)+sum_{k=1}^{infty}dfrac1{(2k-1)^{p+q}}\
    &=O(p)O(q)-s(q, p)+O(p+q)
    \
    end{array}
    $



    Some additional
    definitions and stuff
    which might be useful.



    $Z_{n}(s)
    =sum_{k=1}^n frac1{k^s}
    $



    $E_{n}(s)
    =sum_{k=1}^n frac1{(2k)^s}
    =frac1{2^s}sum_{k=1}^n frac1{k^s}
    =frac1{2^s}Z_{n}(s)
    $



    $O_{n}(s)+E_{n}(s)
    =sum_{k=1}^{2n} frac1{k^s}
    =Z_{2n}(s)
    $
    so
    $O_{n}(s)
    =Z_{2n}(s)-E_{n}(s)
    =Z_{2n}(s)-frac1{2^s}Z_{n}(s)
    $



    $O(s)
    =O_{infty}(s)
    $    ,
    $E(s)
    =E_{infty}(s)
    $    ,
    $Z(s)
    =zeta(s)
    =Z_{infty}(s)
    $    .



    $O(s)
    =Z(s)-E(s)
    =Z(s)-frac1{2^s}Z(s)
    =(1-2^{-s})Z(s)
    $






    share|cite|improve this answer


























      1














      Changing your notation a bit
      and playing around.



      If
      $O_{n}(s)
      =1+frac{1}{3^{s}}+dots+frac{1}{(2n-1)^{s}}
      =sum_{k=1}^n frac1{(2k-1)^s}
      $      ,
      $O(s)
      =O_{infty}(s)
      $      ,
      and
      $s(p, q)
      =sum_{n=1}^{infty}dfrac{O_{n}(p)}{(2n-1)^{q}}
      $
      then



      $s(p, q)+s(q, p)
      =O(p)O(q)+O(p+q)
      $      .



      Proof.



      $begin{array}\
      s(p, q)
      &=sum_{n=1}^{infty}dfrac{O_{n}(p)}{(2n-1)^{q}}\
      &=sum_{n=1}^{infty}dfrac{sum_{k=1}^n frac1{(2k-1)^p}}{(2n-1)^{q}}\
      &=sum_{n=1}^{infty}sum_{k=1}^n dfrac1{(2k-1)^p}dfrac{1}{(2n-1)^{q}}\
      &=sum_{k=1}^{infty}sum_{n=k}^{infty} dfrac1{(2k-1)^p}dfrac{1}{(2n-1)^{q}}\
      &=sum_{k=1}^{infty}dfrac1{(2k-1)^p}sum_{n=k}^{infty} dfrac{1}{(2n-1)^{q}}\
      &=sum_{k=1}^{infty}dfrac1{(2k-1)^p}left(sum_{n=1}^{infty} dfrac{1}{(2n-1)^{q}}-sum_{n=1}^{k-1} dfrac{1}{(2n-1)^{q}}right)\
      &=sum_{k=1}^{infty}dfrac1{(2k-1)^p}left(O(q)-O_{k-1}(q)right)\
      &=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}O_{k-1}(q)\
      &=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}(O_{k}(q)-frac1{(2k-1)^q})\
      &=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}O_{k}(q)+sum_{k=1}^{infty}dfrac1{(2k-1)^p}dfrac1{(2k-1)^q}\
      &=O(p)O(q)-s(q, p)+sum_{k=1}^{infty}dfrac1{(2k-1)^{p+q}}\
      &=O(p)O(q)-s(q, p)+O(p+q)
      \
      end{array}
      $



      Some additional
      definitions and stuff
      which might be useful.



      $Z_{n}(s)
      =sum_{k=1}^n frac1{k^s}
      $



      $E_{n}(s)
      =sum_{k=1}^n frac1{(2k)^s}
      =frac1{2^s}sum_{k=1}^n frac1{k^s}
      =frac1{2^s}Z_{n}(s)
      $



      $O_{n}(s)+E_{n}(s)
      =sum_{k=1}^{2n} frac1{k^s}
      =Z_{2n}(s)
      $
      so
      $O_{n}(s)
      =Z_{2n}(s)-E_{n}(s)
      =Z_{2n}(s)-frac1{2^s}Z_{n}(s)
      $



      $O(s)
      =O_{infty}(s)
      $      ,
      $E(s)
      =E_{infty}(s)
      $      ,
      $Z(s)
      =zeta(s)
      =Z_{infty}(s)
      $      .



      $O(s)
      =Z(s)-E(s)
      =Z(s)-frac1{2^s}Z(s)
      =(1-2^{-s})Z(s)
      $






      share|cite|improve this answer
























        1












        1








        1






        Changing your notation a bit
        and playing around.



        If
        $O_{n}(s)
        =1+frac{1}{3^{s}}+dots+frac{1}{(2n-1)^{s}}
        =sum_{k=1}^n frac1{(2k-1)^s}
        $        ,
        $O(s)
        =O_{infty}(s)
        $        ,
        and
        $s(p, q)
        =sum_{n=1}^{infty}dfrac{O_{n}(p)}{(2n-1)^{q}}
        $
        then



        $s(p, q)+s(q, p)
        =O(p)O(q)+O(p+q)
        $        .



        Proof.



        $begin{array}\
        s(p, q)
        &=sum_{n=1}^{infty}dfrac{O_{n}(p)}{(2n-1)^{q}}\
        &=sum_{n=1}^{infty}dfrac{sum_{k=1}^n frac1{(2k-1)^p}}{(2n-1)^{q}}\
        &=sum_{n=1}^{infty}sum_{k=1}^n dfrac1{(2k-1)^p}dfrac{1}{(2n-1)^{q}}\
        &=sum_{k=1}^{infty}sum_{n=k}^{infty} dfrac1{(2k-1)^p}dfrac{1}{(2n-1)^{q}}\
        &=sum_{k=1}^{infty}dfrac1{(2k-1)^p}sum_{n=k}^{infty} dfrac{1}{(2n-1)^{q}}\
        &=sum_{k=1}^{infty}dfrac1{(2k-1)^p}left(sum_{n=1}^{infty} dfrac{1}{(2n-1)^{q}}-sum_{n=1}^{k-1} dfrac{1}{(2n-1)^{q}}right)\
        &=sum_{k=1}^{infty}dfrac1{(2k-1)^p}left(O(q)-O_{k-1}(q)right)\
        &=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}O_{k-1}(q)\
        &=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}(O_{k}(q)-frac1{(2k-1)^q})\
        &=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}O_{k}(q)+sum_{k=1}^{infty}dfrac1{(2k-1)^p}dfrac1{(2k-1)^q}\
        &=O(p)O(q)-s(q, p)+sum_{k=1}^{infty}dfrac1{(2k-1)^{p+q}}\
        &=O(p)O(q)-s(q, p)+O(p+q)
        \
        end{array}
        $



        Some additional
        definitions and stuff
        which might be useful.



        $Z_{n}(s)
        =sum_{k=1}^n frac1{k^s}
        $



        $E_{n}(s)
        =sum_{k=1}^n frac1{(2k)^s}
        =frac1{2^s}sum_{k=1}^n frac1{k^s}
        =frac1{2^s}Z_{n}(s)
        $



        $O_{n}(s)+E_{n}(s)
        =sum_{k=1}^{2n} frac1{k^s}
        =Z_{2n}(s)
        $
        so
        $O_{n}(s)
        =Z_{2n}(s)-E_{n}(s)
        =Z_{2n}(s)-frac1{2^s}Z_{n}(s)
        $



        $O(s)
        =O_{infty}(s)
        $        ,
        $E(s)
        =E_{infty}(s)
        $        ,
        $Z(s)
        =zeta(s)
        =Z_{infty}(s)
        $        .



        $O(s)
        =Z(s)-E(s)
        =Z(s)-frac1{2^s}Z(s)
        =(1-2^{-s})Z(s)
        $






        share|cite|improve this answer












        Changing your notation a bit
        and playing around.



        If
        $O_{n}(s)
        =1+frac{1}{3^{s}}+dots+frac{1}{(2n-1)^{s}}
        =sum_{k=1}^n frac1{(2k-1)^s}
        $        ,
        $O(s)
        =O_{infty}(s)
        $        ,
        and
        $s(p, q)
        =sum_{n=1}^{infty}dfrac{O_{n}(p)}{(2n-1)^{q}}
        $
        then



        $s(p, q)+s(q, p)
        =O(p)O(q)+O(p+q)
        $        .



        Proof.



        $begin{array}\
        s(p, q)
        &=sum_{n=1}^{infty}dfrac{O_{n}(p)}{(2n-1)^{q}}\
        &=sum_{n=1}^{infty}dfrac{sum_{k=1}^n frac1{(2k-1)^p}}{(2n-1)^{q}}\
        &=sum_{n=1}^{infty}sum_{k=1}^n dfrac1{(2k-1)^p}dfrac{1}{(2n-1)^{q}}\
        &=sum_{k=1}^{infty}sum_{n=k}^{infty} dfrac1{(2k-1)^p}dfrac{1}{(2n-1)^{q}}\
        &=sum_{k=1}^{infty}dfrac1{(2k-1)^p}sum_{n=k}^{infty} dfrac{1}{(2n-1)^{q}}\
        &=sum_{k=1}^{infty}dfrac1{(2k-1)^p}left(sum_{n=1}^{infty} dfrac{1}{(2n-1)^{q}}-sum_{n=1}^{k-1} dfrac{1}{(2n-1)^{q}}right)\
        &=sum_{k=1}^{infty}dfrac1{(2k-1)^p}left(O(q)-O_{k-1}(q)right)\
        &=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}O_{k-1}(q)\
        &=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}(O_{k}(q)-frac1{(2k-1)^q})\
        &=O(p)O(q)-sum_{k=1}^{infty}dfrac1{(2k-1)^p}O_{k}(q)+sum_{k=1}^{infty}dfrac1{(2k-1)^p}dfrac1{(2k-1)^q}\
        &=O(p)O(q)-s(q, p)+sum_{k=1}^{infty}dfrac1{(2k-1)^{p+q}}\
        &=O(p)O(q)-s(q, p)+O(p+q)
        \
        end{array}
        $



        Some additional
        definitions and stuff
        which might be useful.



        $Z_{n}(s)
        =sum_{k=1}^n frac1{k^s}
        $



        $E_{n}(s)
        =sum_{k=1}^n frac1{(2k)^s}
        =frac1{2^s}sum_{k=1}^n frac1{k^s}
        =frac1{2^s}Z_{n}(s)
        $



        $O_{n}(s)+E_{n}(s)
        =sum_{k=1}^{2n} frac1{k^s}
        =Z_{2n}(s)
        $
        so
        $O_{n}(s)
        =Z_{2n}(s)-E_{n}(s)
        =Z_{2n}(s)-frac1{2^s}Z_{n}(s)
        $



        $O(s)
        =O_{infty}(s)
        $        ,
        $E(s)
        =E_{infty}(s)
        $        ,
        $Z(s)
        =zeta(s)
        =Z_{infty}(s)
        $        .



        $O(s)
        =Z(s)-E(s)
        =Z(s)-frac1{2^s}Z(s)
        =(1-2^{-s})Z(s)
        $







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        marty cohen

        72.7k549128




        72.7k549128






























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