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












0












$begingroup$


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











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 20:17












  • $begingroup$
    I'm not sure, which theorem exactly do you have in mind?
    $endgroup$
    – Isak
    Jan 4 at 22:35










  • $begingroup$
    I'm most interested in the case $p=2n,$ $q=2n+1$.
    $endgroup$
    – Isak
    Jan 11 at 22:00
















0












$begingroup$


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











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 20:17












  • $begingroup$
    I'm not sure, which theorem exactly do you have in mind?
    $endgroup$
    – Isak
    Jan 4 at 22:35










  • $begingroup$
    I'm most interested in the case $p=2n,$ $q=2n+1$.
    $endgroup$
    – Isak
    Jan 11 at 22:00














0












0








0


0



$begingroup$


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











$endgroup$




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




share|cite|improve this question








edited Jan 3 at 19:50









mrtaurho

6,09271641




6,09271641










asked Jan 3 at 19:46









IsakIsak

255




255












  • $begingroup$
    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.
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 20:17












  • $begingroup$
    I'm not sure, which theorem exactly do you have in mind?
    $endgroup$
    – Isak
    Jan 4 at 22:35










  • $begingroup$
    I'm most interested in the case $p=2n,$ $q=2n+1$.
    $endgroup$
    – Isak
    Jan 11 at 22:00


















  • $begingroup$
    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.
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 20:17












  • $begingroup$
    I'm not sure, which theorem exactly do you have in mind?
    $endgroup$
    – Isak
    Jan 4 at 22:35










  • $begingroup$
    I'm most interested in the case $p=2n,$ $q=2n+1$.
    $endgroup$
    – Isak
    Jan 11 at 22:00
















$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Jan 3 at 20:17






$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Jan 3 at 20:17














$begingroup$
I'm not sure, which theorem exactly do you have in mind?
$endgroup$
– Isak
Jan 4 at 22:35




$begingroup$
I'm not sure, which theorem exactly do you have in mind?
$endgroup$
– Isak
Jan 4 at 22:35












$begingroup$
I'm most interested in the case $p=2n,$ $q=2n+1$.
$endgroup$
– Isak
Jan 11 at 22:00




$begingroup$
I'm most interested in the case $p=2n,$ $q=2n+1$.
$endgroup$
– Isak
Jan 11 at 22:00










1 Answer
1






active

oldest

votes


















1












$begingroup$

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









$endgroup$













  • $begingroup$
    What about $p+q$ being odd?
    $endgroup$
    – Isak
    Jan 8 at 22:37










  • $begingroup$
    I don't see where my answer depends on the parity of p or q.
    $endgroup$
    – marty cohen
    Jan 9 at 22:44










  • $begingroup$
    I was thinking about the general case $sum_{n=1}^{infty}frac{O_{n}^{(p)}}{(2n-1)^{q}}$
    $endgroup$
    – Isak
    Jan 9 at 23:13












  • $begingroup$
    Don't know about that. I was doing my usual playing around and found, to my surprise, the expression with p and q exchanged. So if you know s(p, q), you can get s(q, p).
    $endgroup$
    – marty cohen
    Jan 9 at 23:52












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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









$endgroup$













  • $begingroup$
    What about $p+q$ being odd?
    $endgroup$
    – Isak
    Jan 8 at 22:37










  • $begingroup$
    I don't see where my answer depends on the parity of p or q.
    $endgroup$
    – marty cohen
    Jan 9 at 22:44










  • $begingroup$
    I was thinking about the general case $sum_{n=1}^{infty}frac{O_{n}^{(p)}}{(2n-1)^{q}}$
    $endgroup$
    – Isak
    Jan 9 at 23:13












  • $begingroup$
    Don't know about that. I was doing my usual playing around and found, to my surprise, the expression with p and q exchanged. So if you know s(p, q), you can get s(q, p).
    $endgroup$
    – marty cohen
    Jan 9 at 23:52
















1












$begingroup$

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









$endgroup$













  • $begingroup$
    What about $p+q$ being odd?
    $endgroup$
    – Isak
    Jan 8 at 22:37










  • $begingroup$
    I don't see where my answer depends on the parity of p or q.
    $endgroup$
    – marty cohen
    Jan 9 at 22:44










  • $begingroup$
    I was thinking about the general case $sum_{n=1}^{infty}frac{O_{n}^{(p)}}{(2n-1)^{q}}$
    $endgroup$
    – Isak
    Jan 9 at 23:13












  • $begingroup$
    Don't know about that. I was doing my usual playing around and found, to my surprise, the expression with p and q exchanged. So if you know s(p, q), you can get s(q, p).
    $endgroup$
    – marty cohen
    Jan 9 at 23:52














1












1








1





$begingroup$

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









$endgroup$



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 Jan 3 at 20:43









marty cohenmarty cohen

75k549130




75k549130












  • $begingroup$
    What about $p+q$ being odd?
    $endgroup$
    – Isak
    Jan 8 at 22:37










  • $begingroup$
    I don't see where my answer depends on the parity of p or q.
    $endgroup$
    – marty cohen
    Jan 9 at 22:44










  • $begingroup$
    I was thinking about the general case $sum_{n=1}^{infty}frac{O_{n}^{(p)}}{(2n-1)^{q}}$
    $endgroup$
    – Isak
    Jan 9 at 23:13












  • $begingroup$
    Don't know about that. I was doing my usual playing around and found, to my surprise, the expression with p and q exchanged. So if you know s(p, q), you can get s(q, p).
    $endgroup$
    – marty cohen
    Jan 9 at 23:52


















  • $begingroup$
    What about $p+q$ being odd?
    $endgroup$
    – Isak
    Jan 8 at 22:37










  • $begingroup$
    I don't see where my answer depends on the parity of p or q.
    $endgroup$
    – marty cohen
    Jan 9 at 22:44










  • $begingroup$
    I was thinking about the general case $sum_{n=1}^{infty}frac{O_{n}^{(p)}}{(2n-1)^{q}}$
    $endgroup$
    – Isak
    Jan 9 at 23:13












  • $begingroup$
    Don't know about that. I was doing my usual playing around and found, to my surprise, the expression with p and q exchanged. So if you know s(p, q), you can get s(q, p).
    $endgroup$
    – marty cohen
    Jan 9 at 23:52
















$begingroup$
What about $p+q$ being odd?
$endgroup$
– Isak
Jan 8 at 22:37




$begingroup$
What about $p+q$ being odd?
$endgroup$
– Isak
Jan 8 at 22:37












$begingroup$
I don't see where my answer depends on the parity of p or q.
$endgroup$
– marty cohen
Jan 9 at 22:44




$begingroup$
I don't see where my answer depends on the parity of p or q.
$endgroup$
– marty cohen
Jan 9 at 22:44












$begingroup$
I was thinking about the general case $sum_{n=1}^{infty}frac{O_{n}^{(p)}}{(2n-1)^{q}}$
$endgroup$
– Isak
Jan 9 at 23:13






$begingroup$
I was thinking about the general case $sum_{n=1}^{infty}frac{O_{n}^{(p)}}{(2n-1)^{q}}$
$endgroup$
– Isak
Jan 9 at 23:13














$begingroup$
Don't know about that. I was doing my usual playing around and found, to my surprise, the expression with p and q exchanged. So if you know s(p, q), you can get s(q, p).
$endgroup$
– marty cohen
Jan 9 at 23:52




$begingroup$
Don't know about that. I was doing my usual playing around and found, to my surprise, the expression with p and q exchanged. So if you know s(p, q), you can get s(q, p).
$endgroup$
– marty cohen
Jan 9 at 23:52


















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