Generalizing...












2












$begingroup$


I am trying to generalize the fact that, for $k>frac12$,




$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$




To reach this I start off with the Fourier series



$$
t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt),qquad |t|leqpi\
sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt)=frac{t^2}4-frac{pi^2}{12}
$$



integrate both sides from $0$ to $x$:
$$
sum_{ngeq1}frac{(-1)^n}{n^2}int_0^xcos(nt)dt=frac{x^3}{12}-frac{pi^2x}{12}\
sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}
$$
plugging in $x=m^{-2k}$ for $mgeq1$, and $k>1/2$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12m^{6k}}-frac{pi^2}{12m^{2k}}$$
Then applying $sumlimits_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$
With the same process, we have



$$begin{align}
sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^5}sin(n/m^{2k})&=frac{pi^2}{72}zeta(6k)-frac1{240}zeta(10k)-frac{7pi^4}{720}zeta(2k)\
end{align}$$
I am trying to find a general form in terms of $zeta$ values of
$$begin{align}
S_j(k)&=sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^j}sin(n/m^{2k}),qquad text{j is odd},quad j>0\
end{align}$$
And as you've seen, I've found up to $j=5$, but I would like to know if a general form exists. Any help is appreciated.






real-analysis sequences-and-series fourier-series riemann-zeta







share|cite|improve this question











$endgroup$












  • $begingroup$
    Sorry, what does $zeta$ stand for?
    $endgroup$
    – Omojola Micheal
    Jan 1 at 11:08










  • $begingroup$
    @Mike I assume that this should be the Riemann Zeta Function $zeta(s)$ hence you can see how the $frac1{4m^{4k}}$ term of the RHS within the line $$sum_{ngeq1}frac{(-1)^n}{n^2}cos(n/m^{2k})=frac1{4m^{4k}}-frac{pi^2}{12}$$ becomes $frac14zeta(4k)$ after summing over all integer $mgeq 1$ which equals the defintion of the Riemann Zeta Function for $operatorname{Re}(4k)>1$.
    $endgroup$
    – mrtaurho
    Jan 1 at 11:56








  • 1




    $begingroup$
    @Did Why did you only changed it in the first line at not beneath where the same notation is used seven more times?
    $endgroup$
    – mrtaurho
    Jan 1 at 12:25






  • 1




    $begingroup$
    @clathratus: Thanks for that!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 19:48






  • 1




    $begingroup$
    How is $$sum_{mgeq 1}left( frac{1}{4m^{2k}} -frac{pi^2}{12}right) $$ convergent? That is what you are summing, right? The first term gives the zeta term but the other term gives $-infty$ if I'm not mistaken.
    $endgroup$
    – Shashi
    Jan 1 at 22:43


















2












$begingroup$


I am trying to generalize the fact that, for $k>frac12$,




$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$




To reach this I start off with the Fourier series



$$
t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt),qquad |t|leqpi\
sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt)=frac{t^2}4-frac{pi^2}{12}
$$



integrate both sides from $0$ to $x$:
$$
sum_{ngeq1}frac{(-1)^n}{n^2}int_0^xcos(nt)dt=frac{x^3}{12}-frac{pi^2x}{12}\
sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}
$$
plugging in $x=m^{-2k}$ for $mgeq1$, and $k>1/2$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12m^{6k}}-frac{pi^2}{12m^{2k}}$$
Then applying $sumlimits_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$
With the same process, we have



$$begin{align}
sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^5}sin(n/m^{2k})&=frac{pi^2}{72}zeta(6k)-frac1{240}zeta(10k)-frac{7pi^4}{720}zeta(2k)\
end{align}$$
I am trying to find a general form in terms of $zeta$ values of
$$begin{align}
S_j(k)&=sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^j}sin(n/m^{2k}),qquad text{j is odd},quad j>0\
end{align}$$
And as you've seen, I've found up to $j=5$, but I would like to know if a general form exists. Any help is appreciated.






real-analysis sequences-and-series fourier-series riemann-zeta







share|cite|improve this question











$endgroup$












  • $begingroup$
    Sorry, what does $zeta$ stand for?
    $endgroup$
    – Omojola Micheal
    Jan 1 at 11:08










  • $begingroup$
    @Mike I assume that this should be the Riemann Zeta Function $zeta(s)$ hence you can see how the $frac1{4m^{4k}}$ term of the RHS within the line $$sum_{ngeq1}frac{(-1)^n}{n^2}cos(n/m^{2k})=frac1{4m^{4k}}-frac{pi^2}{12}$$ becomes $frac14zeta(4k)$ after summing over all integer $mgeq 1$ which equals the defintion of the Riemann Zeta Function for $operatorname{Re}(4k)>1$.
    $endgroup$
    – mrtaurho
    Jan 1 at 11:56








  • 1




    $begingroup$
    @Did Why did you only changed it in the first line at not beneath where the same notation is used seven more times?
    $endgroup$
    – mrtaurho
    Jan 1 at 12:25






  • 1




    $begingroup$
    @clathratus: Thanks for that!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 19:48






  • 1




    $begingroup$
    How is $$sum_{mgeq 1}left( frac{1}{4m^{2k}} -frac{pi^2}{12}right) $$ convergent? That is what you are summing, right? The first term gives the zeta term but the other term gives $-infty$ if I'm not mistaken.
    $endgroup$
    – Shashi
    Jan 1 at 22:43
















2












2








2


4



$begingroup$


I am trying to generalize the fact that, for $k>frac12$,




$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$




To reach this I start off with the Fourier series



$$
t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt),qquad |t|leqpi\
sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt)=frac{t^2}4-frac{pi^2}{12}
$$



integrate both sides from $0$ to $x$:
$$
sum_{ngeq1}frac{(-1)^n}{n^2}int_0^xcos(nt)dt=frac{x^3}{12}-frac{pi^2x}{12}\
sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}
$$
plugging in $x=m^{-2k}$ for $mgeq1$, and $k>1/2$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12m^{6k}}-frac{pi^2}{12m^{2k}}$$
Then applying $sumlimits_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$
With the same process, we have



$$begin{align}
sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^5}sin(n/m^{2k})&=frac{pi^2}{72}zeta(6k)-frac1{240}zeta(10k)-frac{7pi^4}{720}zeta(2k)\
end{align}$$
I am trying to find a general form in terms of $zeta$ values of
$$begin{align}
S_j(k)&=sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^j}sin(n/m^{2k}),qquad text{j is odd},quad j>0\
end{align}$$
And as you've seen, I've found up to $j=5$, but I would like to know if a general form exists. Any help is appreciated.






real-analysis sequences-and-series fourier-series riemann-zeta







share|cite|improve this question











$endgroup$




I am trying to generalize the fact that, for $k>frac12$,




$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$




To reach this I start off with the Fourier series



$$
t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt),qquad |t|leqpi\
sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt)=frac{t^2}4-frac{pi^2}{12}
$$



integrate both sides from $0$ to $x$:
$$
sum_{ngeq1}frac{(-1)^n}{n^2}int_0^xcos(nt)dt=frac{x^3}{12}-frac{pi^2x}{12}\
sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}
$$
plugging in $x=m^{-2k}$ for $mgeq1$, and $k>1/2$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12m^{6k}}-frac{pi^2}{12m^{2k}}$$
Then applying $sumlimits_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$
With the same process, we have



$$begin{align}
sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^5}sin(n/m^{2k})&=frac{pi^2}{72}zeta(6k)-frac1{240}zeta(10k)-frac{7pi^4}{720}zeta(2k)\
end{align}$$
I am trying to find a general form in terms of $zeta$ values of
$$begin{align}
S_j(k)&=sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^j}sin(n/m^{2k}),qquad text{j is odd},quad j>0\
end{align}$$
And as you've seen, I've found up to $j=5$, but I would like to know if a general form exists. Any help is appreciated.






real-analysis sequences-and-series fourier-series riemann-zeta




real-analysis sequences-and-series fourier-series riemann-zeta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 2:06







clathratus

















asked Jan 1 at 10:30









clathratusclathratus

3,713332




3,713332












  • $begingroup$
    Sorry, what does $zeta$ stand for?
    $endgroup$
    – Omojola Micheal
    Jan 1 at 11:08










  • $begingroup$
    @Mike I assume that this should be the Riemann Zeta Function $zeta(s)$ hence you can see how the $frac1{4m^{4k}}$ term of the RHS within the line $$sum_{ngeq1}frac{(-1)^n}{n^2}cos(n/m^{2k})=frac1{4m^{4k}}-frac{pi^2}{12}$$ becomes $frac14zeta(4k)$ after summing over all integer $mgeq 1$ which equals the defintion of the Riemann Zeta Function for $operatorname{Re}(4k)>1$.
    $endgroup$
    – mrtaurho
    Jan 1 at 11:56








  • 1




    $begingroup$
    @Did Why did you only changed it in the first line at not beneath where the same notation is used seven more times?
    $endgroup$
    – mrtaurho
    Jan 1 at 12:25






  • 1




    $begingroup$
    @clathratus: Thanks for that!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 19:48






  • 1




    $begingroup$
    How is $$sum_{mgeq 1}left( frac{1}{4m^{2k}} -frac{pi^2}{12}right) $$ convergent? That is what you are summing, right? The first term gives the zeta term but the other term gives $-infty$ if I'm not mistaken.
    $endgroup$
    – Shashi
    Jan 1 at 22:43




















  • $begingroup$
    Sorry, what does $zeta$ stand for?
    $endgroup$
    – Omojola Micheal
    Jan 1 at 11:08










  • $begingroup$
    @Mike I assume that this should be the Riemann Zeta Function $zeta(s)$ hence you can see how the $frac1{4m^{4k}}$ term of the RHS within the line $$sum_{ngeq1}frac{(-1)^n}{n^2}cos(n/m^{2k})=frac1{4m^{4k}}-frac{pi^2}{12}$$ becomes $frac14zeta(4k)$ after summing over all integer $mgeq 1$ which equals the defintion of the Riemann Zeta Function for $operatorname{Re}(4k)>1$.
    $endgroup$
    – mrtaurho
    Jan 1 at 11:56








  • 1




    $begingroup$
    @Did Why did you only changed it in the first line at not beneath where the same notation is used seven more times?
    $endgroup$
    – mrtaurho
    Jan 1 at 12:25






  • 1




    $begingroup$
    @clathratus: Thanks for that!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 19:48






  • 1




    $begingroup$
    How is $$sum_{mgeq 1}left( frac{1}{4m^{2k}} -frac{pi^2}{12}right) $$ convergent? That is what you are summing, right? The first term gives the zeta term but the other term gives $-infty$ if I'm not mistaken.
    $endgroup$
    – Shashi
    Jan 1 at 22:43


















$begingroup$
Sorry, what does $zeta$ stand for?
$endgroup$
– Omojola Micheal
Jan 1 at 11:08




$begingroup$
Sorry, what does $zeta$ stand for?
$endgroup$
– Omojola Micheal
Jan 1 at 11:08












$begingroup$
@Mike I assume that this should be the Riemann Zeta Function $zeta(s)$ hence you can see how the $frac1{4m^{4k}}$ term of the RHS within the line $$sum_{ngeq1}frac{(-1)^n}{n^2}cos(n/m^{2k})=frac1{4m^{4k}}-frac{pi^2}{12}$$ becomes $frac14zeta(4k)$ after summing over all integer $mgeq 1$ which equals the defintion of the Riemann Zeta Function for $operatorname{Re}(4k)>1$.
$endgroup$
– mrtaurho
Jan 1 at 11:56






$begingroup$
@Mike I assume that this should be the Riemann Zeta Function $zeta(s)$ hence you can see how the $frac1{4m^{4k}}$ term of the RHS within the line $$sum_{ngeq1}frac{(-1)^n}{n^2}cos(n/m^{2k})=frac1{4m^{4k}}-frac{pi^2}{12}$$ becomes $frac14zeta(4k)$ after summing over all integer $mgeq 1$ which equals the defintion of the Riemann Zeta Function for $operatorname{Re}(4k)>1$.
$endgroup$
– mrtaurho
Jan 1 at 11:56






1




1




$begingroup$
@Did Why did you only changed it in the first line at not beneath where the same notation is used seven more times?
$endgroup$
– mrtaurho
Jan 1 at 12:25




$begingroup$
@Did Why did you only changed it in the first line at not beneath where the same notation is used seven more times?
$endgroup$
– mrtaurho
Jan 1 at 12:25




1




1




$begingroup$
@clathratus: Thanks for that!
$endgroup$
– Omojola Micheal
Jan 1 at 19:48




$begingroup$
@clathratus: Thanks for that!
$endgroup$
– Omojola Micheal
Jan 1 at 19:48




1




1




$begingroup$
How is $$sum_{mgeq 1}left( frac{1}{4m^{2k}} -frac{pi^2}{12}right) $$ convergent? That is what you are summing, right? The first term gives the zeta term but the other term gives $-infty$ if I'm not mistaken.
$endgroup$
– Shashi
Jan 1 at 22:43






$begingroup$
How is $$sum_{mgeq 1}left( frac{1}{4m^{2k}} -frac{pi^2}{12}right) $$ convergent? That is what you are summing, right? The first term gives the zeta term but the other term gives $-infty$ if I'm not mistaken.
$endgroup$
– Shashi
Jan 1 at 22:43












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