Formula for harmonic progression $sum _{k=1}^n frac{1}{a k+b}$.
up vote
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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
edited Nov 28 at 15:40
Asaf Karagila♦
300k32422752
|
show 5 more comments
up vote
18
down vote
favorite
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
edited Nov 28 at 15:40
Asaf Karagila♦
300k32422752
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
|
show 5 more comments
up vote
18
down vote
favorite
up vote
18
down vote
favorite
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
edited Nov 28 at 15:40
Asaf Karagila♦
300k32422752
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
edited Nov 28 at 15:40
Asaf Karagila♦
300k32422752
edited Nov 28 at 15:40
Asaf Karagila♦
300k32422752
edited Nov 28 at 15:40
Asaf Karagila♦
300k32422752
edited Nov 28 at 15:40
Asaf Karagila♦
300k32422752
edited Nov 28 at 15:40
Asaf Karagila♦
300k32422752
300k32422752
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
|
show 5 more comments
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
|
show 5 more comments
1 Answer
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accepted
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
add a comment |
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down vote
accepted
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
add a comment |
up vote
4
down vote
accepted
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
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
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
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
edited Nov 29 at 1:38
answered Nov 26 at 23:33
user618463
add a comment |
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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