Formula for harmonic progression $sum _{k=1}^n frac{1}{a k+b}$.












18














Can anyone prove this formula for the harmonic progression easily?




For $a$ and $b$ integers:
begin{multline}sum _{k=1}^n frac{1}{a k+b}=-frac{1}{2b}+frac{1}{2(a n+b)}\+int_0^1 2pi (1-u)sin[a npi u]sin[(a n+2b)pi u]cot[api u],duend{multline}




PS This formula was created by me, am just looking for an indirect proof to show its validity (not sure if it's a good idea to divulge it before publishing the result but anyway, perhaps being here already shows I am the creator).



Now, even more surprising are the patterns of the higher orders:




$$sum_{k=1}^n frac{1}{(a k+b)^3}=-frac{1}{2 b^3}+frac{1}{2(a n+b)^3}\-frac{4pi ^3}{3}int _0^1 left(-u+u^3right) sin{[a npi(1-u)]}sin{[(a n+2b)pi(1-u)]}cot[api (1-u)]du$$




The general formula for odd powers is:




$$sum_{j=1}^{n}frac{1}{(a j+b)^{2k+1}}=-frac{1}{2b^{2k+1}}+frac{1}{2(a n+b)^{2k+1}}\+(-1)^{k}(2pi)^{2k+1}int_{0}^{1}sum_{j=0}^{k}frac{B_{2k-2j}left(2-2^{2k-2j}right)}{(2k-2j)!(2j+1)!}u^{2j+1}f(u,n),du$$




where
$f(u,n)=sin{[a npi(1-u)]}sin{[(a n+2b)pi(1-u)]}cot[api (1-u)]$,



and the general formula for even powers is:




$$sum_{j=1}^{n}frac{1}{(a j+b)^{2k}}=-frac{1}{2b^{2k}}+frac{1}{2(a n+b)^{2k}}\-(-1)^{k}(2pi)^{2k}int_{0}^{1}sum_{j=0}^{k}frac{B_{2k-2j}left(2-2^{2k-2j}right)}{(2k-2j)!(2j)!}u^{2j}g(u,n),du$$




where
$g(u,n)=sin{[a npi(1-u)]}cos{[(a n+2b)pi(1-u)]}cot[api (1-u)]$



These formulae work for the harmonic numbers too ($a=1,b=0$), if term $-1/(2b^k)$ is discarded.






sequences-and-series summation







share|cite|improve this question




















  • 2




    If anyone wants to have at go—pull the sine out of the cotangent, use the geometric series formulae to expand the factor like $tfrac{sin mx/2}{sin x/2}$ into a Fourier series, use the product-to-sum formula for trig functions twice, and integrate by parts. I expect that'll turn the integral into a generalized harmonic series.
    – K B Dave
    Nov 22 at 6:43






  • 1




    You are welcome dear :)
    – Nosrati
    Nov 23 at 7:28






  • 1




    Some easy algebra with the digamma function $psi$ gives that the sum can be written $frac{1}{a}left[psileft(frac{b}{a} + n + 1right) - psileft(frac{b}{a} + 1right)right]$, so one could just as well view this problem as showing an integral identity for $psi$. NB there are some nice integral representations of $psi$, so perhaps one of those could be manipulated to show this one.
    – Travis
    Nov 23 at 9:44






  • 1




    In accordance with the comment of Travis, for $mgeq 2$ you can use the Hurwitz zeta function : $enspacedisplaystylesumlimits_{k=1}^nfrac{1}{(ak+b)^m}=frac{1}{a^m}left(zeta(m,1+frac{b}{a})-zeta(m,1+frac{b}{a}+n)right)$ . Maybe there are some integrals which will help you.
    – user90369
    Nov 23 at 10:04








  • 1




    And if I would like to proof such formulas, I would simply substitute $sin x$ by $frac{e^{ix}-e^{-ix}}{i2}$ and $cot x$ by $ifrac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}}$ . Then check what comes out. :)
    – user90369
    Nov 23 at 10:15


















18














Can anyone prove this formula for the harmonic progression easily?




For $a$ and $b$ integers:
begin{multline}sum _{k=1}^n frac{1}{a k+b}=-frac{1}{2b}+frac{1}{2(a n+b)}\+int_0^1 2pi (1-u)sin[a npi u]sin[(a n+2b)pi u]cot[api u],duend{multline}




PS This formula was created by me, am just looking for an indirect proof to show its validity (not sure if it's a good idea to divulge it before publishing the result but anyway, perhaps being here already shows I am the creator).



Now, even more surprising are the patterns of the higher orders:




$$sum_{k=1}^n frac{1}{(a k+b)^3}=-frac{1}{2 b^3}+frac{1}{2(a n+b)^3}\-frac{4pi ^3}{3}int _0^1 left(-u+u^3right) sin{[a npi(1-u)]}sin{[(a n+2b)pi(1-u)]}cot[api (1-u)]du$$




The general formula for odd powers is:




$$sum_{j=1}^{n}frac{1}{(a j+b)^{2k+1}}=-frac{1}{2b^{2k+1}}+frac{1}{2(a n+b)^{2k+1}}\+(-1)^{k}(2pi)^{2k+1}int_{0}^{1}sum_{j=0}^{k}frac{B_{2k-2j}left(2-2^{2k-2j}right)}{(2k-2j)!(2j+1)!}u^{2j+1}f(u,n),du$$




where
$f(u,n)=sin{[a npi(1-u)]}sin{[(a n+2b)pi(1-u)]}cot[api (1-u)]$,



and the general formula for even powers is:




$$sum_{j=1}^{n}frac{1}{(a j+b)^{2k}}=-frac{1}{2b^{2k}}+frac{1}{2(a n+b)^{2k}}\-(-1)^{k}(2pi)^{2k}int_{0}^{1}sum_{j=0}^{k}frac{B_{2k-2j}left(2-2^{2k-2j}right)}{(2k-2j)!(2j)!}u^{2j}g(u,n),du$$




where
$g(u,n)=sin{[a npi(1-u)]}cos{[(a n+2b)pi(1-u)]}cot[api (1-u)]$



These formulae work for the harmonic numbers too ($a=1,b=0$), if term $-1/(2b^k)$ is discarded.






sequences-and-series summation







share|cite|improve this question




















  • 2




    If anyone wants to have at go—pull the sine out of the cotangent, use the geometric series formulae to expand the factor like $tfrac{sin mx/2}{sin x/2}$ into a Fourier series, use the product-to-sum formula for trig functions twice, and integrate by parts. I expect that'll turn the integral into a generalized harmonic series.
    – K B Dave
    Nov 22 at 6:43






  • 1




    You are welcome dear :)
    – Nosrati
    Nov 23 at 7:28






  • 1




    Some easy algebra with the digamma function $psi$ gives that the sum can be written $frac{1}{a}left[psileft(frac{b}{a} + n + 1right) - psileft(frac{b}{a} + 1right)right]$, so one could just as well view this problem as showing an integral identity for $psi$. NB there are some nice integral representations of $psi$, so perhaps one of those could be manipulated to show this one.
    – Travis
    Nov 23 at 9:44






  • 1




    In accordance with the comment of Travis, for $mgeq 2$ you can use the Hurwitz zeta function : $enspacedisplaystylesumlimits_{k=1}^nfrac{1}{(ak+b)^m}=frac{1}{a^m}left(zeta(m,1+frac{b}{a})-zeta(m,1+frac{b}{a}+n)right)$ . Maybe there are some integrals which will help you.
    – user90369
    Nov 23 at 10:04








  • 1




    And if I would like to proof such formulas, I would simply substitute $sin x$ by $frac{e^{ix}-e^{-ix}}{i2}$ and $cot x$ by $ifrac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}}$ . Then check what comes out. :)
    – user90369
    Nov 23 at 10:15
















18












18








18


10





Can anyone prove this formula for the harmonic progression easily?




For $a$ and $b$ integers:
begin{multline}sum _{k=1}^n frac{1}{a k+b}=-frac{1}{2b}+frac{1}{2(a n+b)}\+int_0^1 2pi (1-u)sin[a npi u]sin[(a n+2b)pi u]cot[api u],duend{multline}




PS This formula was created by me, am just looking for an indirect proof to show its validity (not sure if it's a good idea to divulge it before publishing the result but anyway, perhaps being here already shows I am the creator).



Now, even more surprising are the patterns of the higher orders:




$$sum_{k=1}^n frac{1}{(a k+b)^3}=-frac{1}{2 b^3}+frac{1}{2(a n+b)^3}\-frac{4pi ^3}{3}int _0^1 left(-u+u^3right) sin{[a npi(1-u)]}sin{[(a n+2b)pi(1-u)]}cot[api (1-u)]du$$




The general formula for odd powers is:




$$sum_{j=1}^{n}frac{1}{(a j+b)^{2k+1}}=-frac{1}{2b^{2k+1}}+frac{1}{2(a n+b)^{2k+1}}\+(-1)^{k}(2pi)^{2k+1}int_{0}^{1}sum_{j=0}^{k}frac{B_{2k-2j}left(2-2^{2k-2j}right)}{(2k-2j)!(2j+1)!}u^{2j+1}f(u,n),du$$




where
$f(u,n)=sin{[a npi(1-u)]}sin{[(a n+2b)pi(1-u)]}cot[api (1-u)]$,



and the general formula for even powers is:




$$sum_{j=1}^{n}frac{1}{(a j+b)^{2k}}=-frac{1}{2b^{2k}}+frac{1}{2(a n+b)^{2k}}\-(-1)^{k}(2pi)^{2k}int_{0}^{1}sum_{j=0}^{k}frac{B_{2k-2j}left(2-2^{2k-2j}right)}{(2k-2j)!(2j)!}u^{2j}g(u,n),du$$




where
$g(u,n)=sin{[a npi(1-u)]}cos{[(a n+2b)pi(1-u)]}cot[api (1-u)]$



These formulae work for the harmonic numbers too ($a=1,b=0$), if term $-1/(2b^k)$ is discarded.






sequences-and-series summation







share|cite|improve this question















Can anyone prove this formula for the harmonic progression easily?




For $a$ and $b$ integers:
begin{multline}sum _{k=1}^n frac{1}{a k+b}=-frac{1}{2b}+frac{1}{2(a n+b)}\+int_0^1 2pi (1-u)sin[a npi u]sin[(a n+2b)pi u]cot[api u],duend{multline}




PS This formula was created by me, am just looking for an indirect proof to show its validity (not sure if it's a good idea to divulge it before publishing the result but anyway, perhaps being here already shows I am the creator).



Now, even more surprising are the patterns of the higher orders:




$$sum_{k=1}^n frac{1}{(a k+b)^3}=-frac{1}{2 b^3}+frac{1}{2(a n+b)^3}\-frac{4pi ^3}{3}int _0^1 left(-u+u^3right) sin{[a npi(1-u)]}sin{[(a n+2b)pi(1-u)]}cot[api (1-u)]du$$




The general formula for odd powers is:




$$sum_{j=1}^{n}frac{1}{(a j+b)^{2k+1}}=-frac{1}{2b^{2k+1}}+frac{1}{2(a n+b)^{2k+1}}\+(-1)^{k}(2pi)^{2k+1}int_{0}^{1}sum_{j=0}^{k}frac{B_{2k-2j}left(2-2^{2k-2j}right)}{(2k-2j)!(2j+1)!}u^{2j+1}f(u,n),du$$




where
$f(u,n)=sin{[a npi(1-u)]}sin{[(a n+2b)pi(1-u)]}cot[api (1-u)]$,



and the general formula for even powers is:




$$sum_{j=1}^{n}frac{1}{(a j+b)^{2k}}=-frac{1}{2b^{2k}}+frac{1}{2(a n+b)^{2k}}\-(-1)^{k}(2pi)^{2k}int_{0}^{1}sum_{j=0}^{k}frac{B_{2k-2j}left(2-2^{2k-2j}right)}{(2k-2j)!(2j)!}u^{2j}g(u,n),du$$




where
$g(u,n)=sin{[a npi(1-u)]}cos{[(a n+2b)pi(1-u)]}cot[api (1-u)]$



These formulae work for the harmonic numbers too ($a=1,b=0$), if term $-1/(2b^k)$ is discarded.






sequences-and-series summation




sequences-and-series summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 at 15:40









Asaf Karagila♦

301k32422754




301k32422754










asked Nov 22 at 5:33







user618463















  • 2




    If anyone wants to have at go—pull the sine out of the cotangent, use the geometric series formulae to expand the factor like $tfrac{sin mx/2}{sin x/2}$ into a Fourier series, use the product-to-sum formula for trig functions twice, and integrate by parts. I expect that'll turn the integral into a generalized harmonic series.
    – K B Dave
    Nov 22 at 6:43






  • 1




    You are welcome dear :)
    – Nosrati
    Nov 23 at 7:28






  • 1




    Some easy algebra with the digamma function $psi$ gives that the sum can be written $frac{1}{a}left[psileft(frac{b}{a} + n + 1right) - psileft(frac{b}{a} + 1right)right]$, so one could just as well view this problem as showing an integral identity for $psi$. NB there are some nice integral representations of $psi$, so perhaps one of those could be manipulated to show this one.
    – Travis
    Nov 23 at 9:44






  • 1




    In accordance with the comment of Travis, for $mgeq 2$ you can use the Hurwitz zeta function : $enspacedisplaystylesumlimits_{k=1}^nfrac{1}{(ak+b)^m}=frac{1}{a^m}left(zeta(m,1+frac{b}{a})-zeta(m,1+frac{b}{a}+n)right)$ . Maybe there are some integrals which will help you.
    – user90369
    Nov 23 at 10:04








  • 1




    And if I would like to proof such formulas, I would simply substitute $sin x$ by $frac{e^{ix}-e^{-ix}}{i2}$ and $cot x$ by $ifrac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}}$ . Then check what comes out. :)
    – user90369
    Nov 23 at 10:15
















  • 2




    If anyone wants to have at go—pull the sine out of the cotangent, use the geometric series formulae to expand the factor like $tfrac{sin mx/2}{sin x/2}$ into a Fourier series, use the product-to-sum formula for trig functions twice, and integrate by parts. I expect that'll turn the integral into a generalized harmonic series.
    – K B Dave
    Nov 22 at 6:43






  • 1




    You are welcome dear :)
    – Nosrati
    Nov 23 at 7:28






  • 1




    Some easy algebra with the digamma function $psi$ gives that the sum can be written $frac{1}{a}left[psileft(frac{b}{a} + n + 1right) - psileft(frac{b}{a} + 1right)right]$, so one could just as well view this problem as showing an integral identity for $psi$. NB there are some nice integral representations of $psi$, so perhaps one of those could be manipulated to show this one.
    – Travis
    Nov 23 at 9:44






  • 1




    In accordance with the comment of Travis, for $mgeq 2$ you can use the Hurwitz zeta function : $enspacedisplaystylesumlimits_{k=1}^nfrac{1}{(ak+b)^m}=frac{1}{a^m}left(zeta(m,1+frac{b}{a})-zeta(m,1+frac{b}{a}+n)right)$ . Maybe there are some integrals which will help you.
    – user90369
    Nov 23 at 10:04








  • 1




    And if I would like to proof such formulas, I would simply substitute $sin x$ by $frac{e^{ix}-e^{-ix}}{i2}$ and $cot x$ by $ifrac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}}$ . Then check what comes out. :)
    – user90369
    Nov 23 at 10:15










2




2




If anyone wants to have at go—pull the sine out of the cotangent, use the geometric series formulae to expand the factor like $tfrac{sin mx/2}{sin x/2}$ into a Fourier series, use the product-to-sum formula for trig functions twice, and integrate by parts. I expect that'll turn the integral into a generalized harmonic series.
– K B Dave
Nov 22 at 6:43




If anyone wants to have at go—pull the sine out of the cotangent, use the geometric series formulae to expand the factor like $tfrac{sin mx/2}{sin x/2}$ into a Fourier series, use the product-to-sum formula for trig functions twice, and integrate by parts. I expect that'll turn the integral into a generalized harmonic series.
– K B Dave
Nov 22 at 6:43




1




1




You are welcome dear :)
– Nosrati
Nov 23 at 7:28




You are welcome dear :)
– Nosrati
Nov 23 at 7:28




1




1




Some easy algebra with the digamma function $psi$ gives that the sum can be written $frac{1}{a}left[psileft(frac{b}{a} + n + 1right) - psileft(frac{b}{a} + 1right)right]$, so one could just as well view this problem as showing an integral identity for $psi$. NB there are some nice integral representations of $psi$, so perhaps one of those could be manipulated to show this one.
– Travis
Nov 23 at 9:44




Some easy algebra with the digamma function $psi$ gives that the sum can be written $frac{1}{a}left[psileft(frac{b}{a} + n + 1right) - psileft(frac{b}{a} + 1right)right]$, so one could just as well view this problem as showing an integral identity for $psi$. NB there are some nice integral representations of $psi$, so perhaps one of those could be manipulated to show this one.
– Travis
Nov 23 at 9:44




1




1




In accordance with the comment of Travis, for $mgeq 2$ you can use the Hurwitz zeta function : $enspacedisplaystylesumlimits_{k=1}^nfrac{1}{(ak+b)^m}=frac{1}{a^m}left(zeta(m,1+frac{b}{a})-zeta(m,1+frac{b}{a}+n)right)$ . Maybe there are some integrals which will help you.
– user90369
Nov 23 at 10:04






In accordance with the comment of Travis, for $mgeq 2$ you can use the Hurwitz zeta function : $enspacedisplaystylesumlimits_{k=1}^nfrac{1}{(ak+b)^m}=frac{1}{a^m}left(zeta(m,1+frac{b}{a})-zeta(m,1+frac{b}{a}+n)right)$ . Maybe there are some integrals which will help you.
– user90369
Nov 23 at 10:04






1




1




And if I would like to proof such formulas, I would simply substitute $sin x$ by $frac{e^{ix}-e^{-ix}}{i2}$ and $cot x$ by $ifrac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}}$ . Then check what comes out. :)
– user90369
Nov 23 at 10:15






And if I would like to proof such formulas, I would simply substitute $sin x$ by $frac{e^{ix}-e^{-ix}}{i2}$ and $cot x$ by $ifrac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}}$ . Then check what comes out. :)
– user90369
Nov 23 at 10:15












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Looks like my post was vandalized.
I'm able to edit this answer though, so let me include a link to the paper on arXiv, which has now been accepted.



https://arxiv.org/abs/1811.11305






share|cite|improve this answer























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    Looks like my post was vandalized.
    I'm able to edit this answer though, so let me include a link to the paper on arXiv, which has now been accepted.



    https://arxiv.org/abs/1811.11305






    share|cite|improve this answer




























      4














      Looks like my post was vandalized.
      I'm able to edit this answer though, so let me include a link to the paper on arXiv, which has now been accepted.



      https://arxiv.org/abs/1811.11305






      share|cite|improve this answer


























        4












        4








        4






        Looks like my post was vandalized.
        I'm able to edit this answer though, so let me include a link to the paper on arXiv, which has now been accepted.



        https://arxiv.org/abs/1811.11305






        share|cite|improve this answer














        Looks like my post was vandalized.
        I'm able to edit this answer though, so let me include a link to the paper on arXiv, which has now been accepted.



        https://arxiv.org/abs/1811.11305







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 29 at 1:38

























        answered Nov 26 at 23:33







        user618463





































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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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