Closed form expression for (periodic) generalized harmonic numbers?
$begingroup$
As far as I understand, there does not exist a pure closed form expression for the generalized harmonic numbers $H_{n,m}=sum_{k=1}^n frac{1}{k^m}$ with $minmathbb R$.
My question is, however, whether there might exist a closed form expression (or maybe nice identities or an asymptotic expansion?) of which I call `periodic generalized harmonic numbers' $P_{n,m,theta}$ in which I define $$P_{n,m,theta}=sum_{k=1}^n frac{cos(thetaln(k))}{k^m}$$ or w.l.o.g. $$P_{n,m,theta}=sum_{k=1}^n frac{sin(thetaln(k))}{k^m}$$ with $thetainmathbb R$.
One could also see this as an expression for the corresponding partial sums.
At the moment, I know that one could derive the Euler-Maclaurin expansions of them but since one obtains a rather nontrivial series containing Bernoulli numbers and the derivatives of the defining function, I don't know what to make from it. Maybe there do exist nice alternatives to the Euler-Maclaurin expansion?
Concludingly, I know that both series are divergent and a linear combination of them gives us the defining series for the Riemann zeta function. See also my slightly related question Equality of perturbed Ramanujan's sum and -1/12 in which a `perturbed' $H_{n,-1}$ approximates $-frac{1}{12}$$=zeta(-1)$.
sequences-and-series trigonometry asymptotics closed-form divergent-series
asked Dec 22 '18 at 16:15
FrozenharpFrozenharp
244
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add a comment |
$begingroup$
As far as I understand, there does not exist a pure closed form expression for the generalized harmonic numbers $H_{n,m}=sum_{k=1}^n frac{1}{k^m}$ with $minmathbb R$.
My question is, however, whether there might exist a closed form expression (or maybe nice identities or an asymptotic expansion?) of which I call `periodic generalized harmonic numbers' $P_{n,m,theta}$ in which I define $$P_{n,m,theta}=sum_{k=1}^n frac{cos(thetaln(k))}{k^m}$$ or w.l.o.g. $$P_{n,m,theta}=sum_{k=1}^n frac{sin(thetaln(k))}{k^m}$$ with $thetainmathbb R$.
One could also see this as an expression for the corresponding partial sums.
At the moment, I know that one could derive the Euler-Maclaurin expansions of them but since one obtains a rather nontrivial series containing Bernoulli numbers and the derivatives of the defining function, I don't know what to make from it. Maybe there do exist nice alternatives to the Euler-Maclaurin expansion?
Concludingly, I know that both series are divergent and a linear combination of them gives us the defining series for the Riemann zeta function. See also my slightly related question Equality of perturbed Ramanujan's sum and -1/12 in which a `perturbed' $H_{n,-1}$ approximates $-frac{1}{12}$$=zeta(-1)$.
sequences-and-series trigonometry asymptotics closed-form divergent-series
asked Dec 22 '18 at 16:15
FrozenharpFrozenharp
244
$endgroup$
$begingroup$
Welcome to the site. This is a good question. I can't answer it, but I know there are certain users here who can. Hope they see it.
$endgroup$
– Antonio Vargas
Dec 23 '18 at 20:45
$begingroup$
Thank you! I hope so: Without giving full disclosure I've found a method which, when operating on the according partial sums expressions (if they do exist of course), might give us necessary conditions for the real part of a zero of the Riemann zeta function (hence might solve the Riemann hypothesis). My thesis is about this matter but I strongly suppose I can't find a partial sum expression by myself; when one can indeed find such an expression and the according method gives more insights in the real part of the zeroes of the zeta function I will let him/her personally know ;).
$endgroup$
– Frozenharp
Dec 23 '18 at 21:19
add a comment |
$begingroup$
As far as I understand, there does not exist a pure closed form expression for the generalized harmonic numbers $H_{n,m}=sum_{k=1}^n frac{1}{k^m}$ with $minmathbb R$.
My question is, however, whether there might exist a closed form expression (or maybe nice identities or an asymptotic expansion?) of which I call `periodic generalized harmonic numbers' $P_{n,m,theta}$ in which I define $$P_{n,m,theta}=sum_{k=1}^n frac{cos(thetaln(k))}{k^m}$$ or w.l.o.g. $$P_{n,m,theta}=sum_{k=1}^n frac{sin(thetaln(k))}{k^m}$$ with $thetainmathbb R$.
One could also see this as an expression for the corresponding partial sums.
At the moment, I know that one could derive the Euler-Maclaurin expansions of them but since one obtains a rather nontrivial series containing Bernoulli numbers and the derivatives of the defining function, I don't know what to make from it. Maybe there do exist nice alternatives to the Euler-Maclaurin expansion?
Concludingly, I know that both series are divergent and a linear combination of them gives us the defining series for the Riemann zeta function. See also my slightly related question Equality of perturbed Ramanujan's sum and -1/12 in which a `perturbed' $H_{n,-1}$ approximates $-frac{1}{12}$$=zeta(-1)$.
sequences-and-series trigonometry asymptotics closed-form divergent-series
asked Dec 22 '18 at 16:15
FrozenharpFrozenharp
244
$endgroup$
As far as I understand, there does not exist a pure closed form expression for the generalized harmonic numbers $H_{n,m}=sum_{k=1}^n frac{1}{k^m}$ with $minmathbb R$.
My question is, however, whether there might exist a closed form expression (or maybe nice identities or an asymptotic expansion?) of which I call `periodic generalized harmonic numbers' $P_{n,m,theta}$ in which I define $$P_{n,m,theta}=sum_{k=1}^n frac{cos(thetaln(k))}{k^m}$$ or w.l.o.g. $$P_{n,m,theta}=sum_{k=1}^n frac{sin(thetaln(k))}{k^m}$$ with $thetainmathbb R$.
One could also see this as an expression for the corresponding partial sums.
At the moment, I know that one could derive the Euler-Maclaurin expansions of them but since one obtains a rather nontrivial series containing Bernoulli numbers and the derivatives of the defining function, I don't know what to make from it. Maybe there do exist nice alternatives to the Euler-Maclaurin expansion?
Concludingly, I know that both series are divergent and a linear combination of them gives us the defining series for the Riemann zeta function. See also my slightly related question Equality of perturbed Ramanujan's sum and -1/12 in which a `perturbed' $H_{n,-1}$ approximates $-frac{1}{12}$$=zeta(-1)$.
sequences-and-series trigonometry asymptotics closed-form divergent-series
sequences-and-series trigonometry asymptotics closed-form divergent-series
asked Dec 22 '18 at 16:15
FrozenharpFrozenharp
244
asked Dec 22 '18 at 16:15
FrozenharpFrozenharp
244
edited Dec 22 '18 at 16:46
Frozenharp
asked Dec 22 '18 at 16:15
FrozenharpFrozenharp
244
asked Dec 22 '18 at 16:15
FrozenharpFrozenharp
244
asked Dec 22 '18 at 16:15
FrozenharpFrozenharp
244
244
$begingroup$
Welcome to the site. This is a good question. I can't answer it, but I know there are certain users here who can. Hope they see it.
$endgroup$
– Antonio Vargas
Dec 23 '18 at 20:45
$begingroup$
Thank you! I hope so: Without giving full disclosure I've found a method which, when operating on the according partial sums expressions (if they do exist of course), might give us necessary conditions for the real part of a zero of the Riemann zeta function (hence might solve the Riemann hypothesis). My thesis is about this matter but I strongly suppose I can't find a partial sum expression by myself; when one can indeed find such an expression and the according method gives more insights in the real part of the zeroes of the zeta function I will let him/her personally know ;).
$endgroup$
– Frozenharp
Dec 23 '18 at 21:19
add a comment |
$begingroup$
Welcome to the site. This is a good question. I can't answer it, but I know there are certain users here who can. Hope they see it.
$endgroup$
– Antonio Vargas
Dec 23 '18 at 20:45
$begingroup$
Thank you! I hope so: Without giving full disclosure I've found a method which, when operating on the according partial sums expressions (if they do exist of course), might give us necessary conditions for the real part of a zero of the Riemann zeta function (hence might solve the Riemann hypothesis). My thesis is about this matter but I strongly suppose I can't find a partial sum expression by myself; when one can indeed find such an expression and the according method gives more insights in the real part of the zeroes of the zeta function I will let him/her personally know ;).
$endgroup$
– Frozenharp
Dec 23 '18 at 21:19
$begingroup$
Welcome to the site. This is a good question. I can't answer it, but I know there are certain users here who can. Hope they see it.
$endgroup$
– Antonio Vargas
Dec 23 '18 at 20:45
$begingroup$
Welcome to the site. This is a good question. I can't answer it, but I know there are certain users here who can. Hope they see it.
$endgroup$
– Antonio Vargas
Dec 23 '18 at 20:45
$begingroup$
Thank you! I hope so: Without giving full disclosure I've found a method which, when operating on the according partial sums expressions (if they do exist of course), might give us necessary conditions for the real part of a zero of the Riemann zeta function (hence might solve the Riemann hypothesis). My thesis is about this matter but I strongly suppose I can't find a partial sum expression by myself; when one can indeed find such an expression and the according method gives more insights in the real part of the zeroes of the zeta function I will let him/her personally know ;).
$endgroup$
– Frozenharp
Dec 23 '18 at 21:19
$begingroup$
Thank you! I hope so: Without giving full disclosure I've found a method which, when operating on the according partial sums expressions (if they do exist of course), might give us necessary conditions for the real part of a zero of the Riemann zeta function (hence might solve the Riemann hypothesis). My thesis is about this matter but I strongly suppose I can't find a partial sum expression by myself; when one can indeed find such an expression and the according method gives more insights in the real part of the zeroes of the zeta function I will let him/her personally know ;).
$endgroup$
– Frozenharp
Dec 23 '18 at 21:19
add a comment |
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$begingroup$
Welcome to the site. This is a good question. I can't answer it, but I know there are certain users here who can. Hope they see it.
$endgroup$
– Antonio Vargas
Dec 23 '18 at 20:45
$begingroup$
Thank you! I hope so: Without giving full disclosure I've found a method which, when operating on the according partial sums expressions (if they do exist of course), might give us necessary conditions for the real part of a zero of the Riemann zeta function (hence might solve the Riemann hypothesis). My thesis is about this matter but I strongly suppose I can't find a partial sum expression by myself; when one can indeed find such an expression and the according method gives more insights in the real part of the zeroes of the zeta function I will let him/her personally know ;).
$endgroup$
– Frozenharp
Dec 23 '18 at 21:19