Mixing
General formula for $sum_{n=0}^infty frac{1}{(an+b)^c}, a> 0, b>0, c>1$
$begingroup$
Is there a general formula for this sum? $$I(a,b,c)=sum_{n=0}^infty frac{1}{(an+b)^c}, a> 0, b>0, c>1$$ I noticed that the first few a, b, and c values yielded seemingly different results, but all look related in some way to the zeta function. Here are a few values of I. $$I(2, 1, 2)= frac{pi^2}{8}$$ $$I(2, 1, 3)= frac{7zeta(3)}{8}$$ $$I(3, 1, 2)= frac{psi^{(1)}(frac{1}{3})}{9}$$ $$I(1,2,2)= frac{pi^2}{6} - 1$$ $$I(pi,pi,pi)= frac{zeta(pi)}{pi^{pi}}$$ Don't pay too much attention to the bounds for a and b as I don't really know if anything changes should the values be negative. Perhaps if a general formula does not exist, maybe one does when fixing certain values such as setting b = 1?
sequences-and-series summation riemann-zeta
asked Jan 8 at 0:12
Suchetan DonthaSuchetan Dontha
14312
$endgroup$
|
show 2 more comments
$begingroup$
Is there a general formula for this sum? $$I(a,b,c)=sum_{n=0}^infty frac{1}{(an+b)^c}, a> 0, b>0, c>1$$ I noticed that the first few a, b, and c values yielded seemingly different results, but all look related in some way to the zeta function. Here are a few values of I. $$I(2, 1, 2)= frac{pi^2}{8}$$ $$I(2, 1, 3)= frac{7zeta(3)}{8}$$ $$I(3, 1, 2)= frac{psi^{(1)}(frac{1}{3})}{9}$$ $$I(1,2,2)= frac{pi^2}{6} - 1$$ $$I(pi,pi,pi)= frac{zeta(pi)}{pi^{pi}}$$ Don't pay too much attention to the bounds for a and b as I don't really know if anything changes should the values be negative. Perhaps if a general formula does not exist, maybe one does when fixing certain values such as setting b = 1?
sequences-and-series summation riemann-zeta
asked Jan 8 at 0:12
Suchetan DonthaSuchetan Dontha
14312
$endgroup$
$begingroup$
no, there isnt a closed formula for that
$endgroup$
– Masacroso
Jan 8 at 0:17
$begingroup$
Some special values can be obtained through the zeta function, e.g. $a=1, b=0$. For example, for $n in Bbb N$ $$zeta (2n)={frac {(-1)^{n+1}B_{2n}(2pi )^{2n}}{2(2n)!}}$$ $${displaystyle zeta (-n)=(-1)^{n}{frac {B_{n+1}}{n+1}}}$$ Other identities can be found at en.wikipedia.org/wiki/Riemann_zeta_function. .
$endgroup$
– Eevee Trainer
Jan 8 at 0:20
$begingroup$
But I imagine you already know the previous going by the tags and some of the values. To my knowledge there is no closed form known for odd integer inputs of the zeta function (maybe part of why Apery's constant, $zeta(3)$, is not known to be transcendental or not). Which, in light of that, (a) might mean this question will likely not yield anything more than partial answers and (b) might be better off on Math Overflow, but that's just a guess. At the very least it would be a very nontrivial question and I'm not the one to ask if things are better on MO.
$endgroup$
– Eevee Trainer
Jan 8 at 0:20
$begingroup$
@EeveeTrainer Thank you for your comment. I've edited the question to be more open and reasonable. Further, I'm not too knowledgable about the zeta function given my high school calculus knowledge. Any information would help me!
$endgroup$
– Suchetan Dontha
Jan 8 at 0:22
$begingroup$
With respect to one particular case ($a=b=1, c=2$), there is a discussion on it in this video - youtube.com/watch?v=aYc9ManKC-s - I haven't watched it yet so I don't have anything to offer there. Just figured you might find it useful.
$endgroup$
– Eevee Trainer
Jan 8 at 0:25
|
show 2 more comments
$begingroup$
Is there a general formula for this sum? $$I(a,b,c)=sum_{n=0}^infty frac{1}{(an+b)^c}, a> 0, b>0, c>1$$ I noticed that the first few a, b, and c values yielded seemingly different results, but all look related in some way to the zeta function. Here are a few values of I. $$I(2, 1, 2)= frac{pi^2}{8}$$ $$I(2, 1, 3)= frac{7zeta(3)}{8}$$ $$I(3, 1, 2)= frac{psi^{(1)}(frac{1}{3})}{9}$$ $$I(1,2,2)= frac{pi^2}{6} - 1$$ $$I(pi,pi,pi)= frac{zeta(pi)}{pi^{pi}}$$ Don't pay too much attention to the bounds for a and b as I don't really know if anything changes should the values be negative. Perhaps if a general formula does not exist, maybe one does when fixing certain values such as setting b = 1?
sequences-and-series summation riemann-zeta
asked Jan 8 at 0:12
Suchetan DonthaSuchetan Dontha
14312
$endgroup$
Is there a general formula for this sum? $$I(a,b,c)=sum_{n=0}^infty frac{1}{(an+b)^c}, a> 0, b>0, c>1$$ I noticed that the first few a, b, and c values yielded seemingly different results, but all look related in some way to the zeta function. Here are a few values of I. $$I(2, 1, 2)= frac{pi^2}{8}$$ $$I(2, 1, 3)= frac{7zeta(3)}{8}$$ $$I(3, 1, 2)= frac{psi^{(1)}(frac{1}{3})}{9}$$ $$I(1,2,2)= frac{pi^2}{6} - 1$$ $$I(pi,pi,pi)= frac{zeta(pi)}{pi^{pi}}$$ Don't pay too much attention to the bounds for a and b as I don't really know if anything changes should the values be negative. Perhaps if a general formula does not exist, maybe one does when fixing certain values such as setting b = 1?
sequences-and-series summation riemann-zeta
sequences-and-series summation riemann-zeta
asked Jan 8 at 0:12
Suchetan DonthaSuchetan Dontha
14312
asked Jan 8 at 0:12
Suchetan DonthaSuchetan Dontha
14312
edited Jan 8 at 0:20
Suchetan Dontha
asked Jan 8 at 0:12
Suchetan DonthaSuchetan Dontha
14312
asked Jan 8 at 0:12
Suchetan DonthaSuchetan Dontha
14312
asked Jan 8 at 0:12
Suchetan DonthaSuchetan Dontha
14312
14312
$begingroup$
no, there isnt a closed formula for that
$endgroup$
– Masacroso
Jan 8 at 0:17
$begingroup$
Some special values can be obtained through the zeta function, e.g. $a=1, b=0$. For example, for $n in Bbb N$ $$zeta (2n)={frac {(-1)^{n+1}B_{2n}(2pi )^{2n}}{2(2n)!}}$$ $${displaystyle zeta (-n)=(-1)^{n}{frac {B_{n+1}}{n+1}}}$$ Other identities can be found at en.wikipedia.org/wiki/Riemann_zeta_function. .
$endgroup$
– Eevee Trainer
Jan 8 at 0:20
$begingroup$
But I imagine you already know the previous going by the tags and some of the values. To my knowledge there is no closed form known for odd integer inputs of the zeta function (maybe part of why Apery's constant, $zeta(3)$, is not known to be transcendental or not). Which, in light of that, (a) might mean this question will likely not yield anything more than partial answers and (b) might be better off on Math Overflow, but that's just a guess. At the very least it would be a very nontrivial question and I'm not the one to ask if things are better on MO.
$endgroup$
– Eevee Trainer
Jan 8 at 0:20
$begingroup$
@EeveeTrainer Thank you for your comment. I've edited the question to be more open and reasonable. Further, I'm not too knowledgable about the zeta function given my high school calculus knowledge. Any information would help me!
$endgroup$
– Suchetan Dontha
Jan 8 at 0:22
$begingroup$
With respect to one particular case ($a=b=1, c=2$), there is a discussion on it in this video - youtube.com/watch?v=aYc9ManKC-s - I haven't watched it yet so I don't have anything to offer there. Just figured you might find it useful.
$endgroup$
– Eevee Trainer
Jan 8 at 0:25
|
show 2 more comments
$begingroup$
no, there isnt a closed formula for that
$endgroup$
– Masacroso
Jan 8 at 0:17
$begingroup$
Some special values can be obtained through the zeta function, e.g. $a=1, b=0$. For example, for $n in Bbb N$ $$zeta (2n)={frac {(-1)^{n+1}B_{2n}(2pi )^{2n}}{2(2n)!}}$$ $${displaystyle zeta (-n)=(-1)^{n}{frac {B_{n+1}}{n+1}}}$$ Other identities can be found at en.wikipedia.org/wiki/Riemann_zeta_function. .
$endgroup$
– Eevee Trainer
Jan 8 at 0:20
$begingroup$
But I imagine you already know the previous going by the tags and some of the values. To my knowledge there is no closed form known for odd integer inputs of the zeta function (maybe part of why Apery's constant, $zeta(3)$, is not known to be transcendental or not). Which, in light of that, (a) might mean this question will likely not yield anything more than partial answers and (b) might be better off on Math Overflow, but that's just a guess. At the very least it would be a very nontrivial question and I'm not the one to ask if things are better on MO.
$endgroup$
– Eevee Trainer
Jan 8 at 0:20
$begingroup$
@EeveeTrainer Thank you for your comment. I've edited the question to be more open and reasonable. Further, I'm not too knowledgable about the zeta function given my high school calculus knowledge. Any information would help me!
$endgroup$
– Suchetan Dontha
Jan 8 at 0:22
$begingroup$
With respect to one particular case ($a=b=1, c=2$), there is a discussion on it in this video - youtube.com/watch?v=aYc9ManKC-s - I haven't watched it yet so I don't have anything to offer there. Just figured you might find it useful.
$endgroup$
– Eevee Trainer
Jan 8 at 0:25
$begingroup$
no, there isnt a closed formula for that
$endgroup$
– Masacroso
Jan 8 at 0:17
$begingroup$
no, there isnt a closed formula for that
$endgroup$
– Masacroso
Jan 8 at 0:17
$begingroup$
Some special values can be obtained through the zeta function, e.g. $a=1, b=0$. For example, for $n in Bbb N$ $$zeta (2n)={frac {(-1)^{n+1}B_{2n}(2pi )^{2n}}{2(2n)!}}$$ $${displaystyle zeta (-n)=(-1)^{n}{frac {B_{n+1}}{n+1}}}$$ Other identities can be found at en.wikipedia.org/wiki/Riemann_zeta_function. .
$endgroup$
– Eevee Trainer
Jan 8 at 0:20
$begingroup$
Some special values can be obtained through the zeta function, e.g. $a=1, b=0$. For example, for $n in Bbb N$ $$zeta (2n)={frac {(-1)^{n+1}B_{2n}(2pi )^{2n}}{2(2n)!}}$$ $${displaystyle zeta (-n)=(-1)^{n}{frac {B_{n+1}}{n+1}}}$$ Other identities can be found at en.wikipedia.org/wiki/Riemann_zeta_function. .
$endgroup$
– Eevee Trainer
Jan 8 at 0:20
$begingroup$
But I imagine you already know the previous going by the tags and some of the values. To my knowledge there is no closed form known for odd integer inputs of the zeta function (maybe part of why Apery's constant, $zeta(3)$, is not known to be transcendental or not). Which, in light of that, (a) might mean this question will likely not yield anything more than partial answers and (b) might be better off on Math Overflow, but that's just a guess. At the very least it would be a very nontrivial question and I'm not the one to ask if things are better on MO.
$endgroup$
– Eevee Trainer
Jan 8 at 0:20
$begingroup$
But I imagine you already know the previous going by the tags and some of the values. To my knowledge there is no closed form known for odd integer inputs of the zeta function (maybe part of why Apery's constant, $zeta(3)$, is not known to be transcendental or not). Which, in light of that, (a) might mean this question will likely not yield anything more than partial answers and (b) might be better off on Math Overflow, but that's just a guess. At the very least it would be a very nontrivial question and I'm not the one to ask if things are better on MO.
$endgroup$
– Eevee Trainer
Jan 8 at 0:20
$begingroup$
@EeveeTrainer Thank you for your comment. I've edited the question to be more open and reasonable. Further, I'm not too knowledgable about the zeta function given my high school calculus knowledge. Any information would help me!
$endgroup$
– Suchetan Dontha
Jan 8 at 0:22
$begingroup$
@EeveeTrainer Thank you for your comment. I've edited the question to be more open and reasonable. Further, I'm not too knowledgable about the zeta function given my high school calculus knowledge. Any information would help me!
$endgroup$
– Suchetan Dontha
Jan 8 at 0:22
$begingroup$
With respect to one particular case ($a=b=1, c=2$), there is a discussion on it in this video - youtube.com/watch?v=aYc9ManKC-s - I haven't watched it yet so I don't have anything to offer there. Just figured you might find it useful.
$endgroup$
– Eevee Trainer
Jan 8 at 0:25
$begingroup$
With respect to one particular case ($a=b=1, c=2$), there is a discussion on it in this video - youtube.com/watch?v=aYc9ManKC-s - I haven't watched it yet so I don't have anything to offer there. Just figured you might find it useful.
$endgroup$
– Eevee Trainer
Jan 8 at 0:25
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
$I(a,b,c)
=sum_{n=0}^infty frac{1}{(an+b)^c}
$
Some simple cases.
$I(1,1,c)
=sum_{n=0}^infty frac{1}{(n+1)^c}
=sum_{n=1}^infty frac{1}{n^c}
=zeta(c)
$
$I(2,2,c)
=sum_{n=0}^infty frac{1}{(2n+2)^c}
=2^{-c}sum_{n=1}^infty frac{1}{n^c}
=2^{-c}zeta(c)
$
$begin{array}\
I(2,1,c)
&=sum_{n=0}^infty frac{1}{(2n+1)^c}\
&=sum_{n=0}^infty frac{1}{(2n+1)^c}+sum_{n=0}^infty frac{1}{(2n+2)^c}-sum_{n=0}^infty frac{1}{(2n+2)^c}\
&=sum_{n=1}^infty frac{1}{n^c}-I(2, 2, c)\
&=I(1, 1, c)-I(2, 2, c)\
&=zeta(c)-2^{-c}zeta(c)\
&=(1-2^{-c})zeta(c)\
end{array}
$
If $b$ is an integer,
$I(1,b,c)
=sum_{n=0}^infty frac{1}{(n+b)^c}
=sum_{n=b}^infty frac{1}{n^c}
=sum_{n=1}^infty frac{1}{n^c}-sum_{n=1}^{b-1} frac{1}{n^c}
=zeta(c)-sum_{n=1}^{b-1} frac{1}{n^c}
$
Note that
$I(a,b,c)
=sum_{n=0}^infty frac{1}{(an+b)^c}
=a^{-c}sum_{n=0}^infty frac{1}{(n+b/a)^c}
=a^{-c}zeta(c, b/a)
$
the Hurwitz zeta function.
See https://en.wikipedia.org/wiki/Hurwitz_zeta_function
for a discussion of this.
answered Jan 8 at 0:29
marty cohenmarty cohen
73.3k549128
$endgroup$
$begingroup$
Thanks for your answer! This was exactly what I was looking for when I first thought of the question. I can't wait to study this function more.
$endgroup$
– Suchetan Dontha
Jan 8 at 0:34
$begingroup$
See my addition at the end of my answer.
$endgroup$
– marty cohen
Jan 8 at 0:43
add a comment |
$begingroup$
While not a complete answer, if $c$ is a positive integer greater than 1 the sum can be written in terms of the polygamma function $psi^{(m)}(z)$.
From the series representation for the polygamma function, namely
$$psi^{(m)}(z) = (-1)^{m + 1} Gamma (m + 1) sum_{n = 0}^infty frac{1}{(n + z)^{m + 1}}, quad m > 0, z notin 0,mathbb{Z}^-$$
if we rewrite your sum as
$$I(a,b,c) = frac{1}{a^c} sum_{n = 0}^infty frac{1}{(n + b/a)^{(c - 1) + 1}},$$
in terms of the polygamma function the sum becomes
$$I(a,b,c) = frac{(-1)^c}{a^c Gamma (c)} psi^{(c - 1)} left (frac{b}{a} right ).$$
Here $c > 1$ where $c in mathbb{N}$, $a > 0$, and $b neq 0,-a,-2a,ldots$
answered Jan 8 at 1:36
omegadotomegadot
5,4422728
$endgroup$
add a comment |
$begingroup$
For $a=b=c=k$, where k is a constant, it seems as though $I(k, k, k) = frac{zeta(k)}{k^k}$
answered Jan 8 at 0:54
Suchetan DonthaSuchetan Dontha
14312
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$I(a,b,c)
=sum_{n=0}^infty frac{1}{(an+b)^c}
$
Some simple cases.
$I(1,1,c)
=sum_{n=0}^infty frac{1}{(n+1)^c}
=sum_{n=1}^infty frac{1}{n^c}
=zeta(c)
$
$I(2,2,c)
=sum_{n=0}^infty frac{1}{(2n+2)^c}
=2^{-c}sum_{n=1}^infty frac{1}{n^c}
=2^{-c}zeta(c)
$
$begin{array}\
I(2,1,c)
&=sum_{n=0}^infty frac{1}{(2n+1)^c}\
&=sum_{n=0}^infty frac{1}{(2n+1)^c}+sum_{n=0}^infty frac{1}{(2n+2)^c}-sum_{n=0}^infty frac{1}{(2n+2)^c}\
&=sum_{n=1}^infty frac{1}{n^c}-I(2, 2, c)\
&=I(1, 1, c)-I(2, 2, c)\
&=zeta(c)-2^{-c}zeta(c)\
&=(1-2^{-c})zeta(c)\
end{array}
$
If $b$ is an integer,
$I(1,b,c)
=sum_{n=0}^infty frac{1}{(n+b)^c}
=sum_{n=b}^infty frac{1}{n^c}
=sum_{n=1}^infty frac{1}{n^c}-sum_{n=1}^{b-1} frac{1}{n^c}
=zeta(c)-sum_{n=1}^{b-1} frac{1}{n^c}
$
Note that
$I(a,b,c)
=sum_{n=0}^infty frac{1}{(an+b)^c}
=a^{-c}sum_{n=0}^infty frac{1}{(n+b/a)^c}
=a^{-c}zeta(c, b/a)
$
the Hurwitz zeta function.
See https://en.wikipedia.org/wiki/Hurwitz_zeta_function
for a discussion of this.
answered Jan 8 at 0:29
marty cohenmarty cohen
73.3k549128
$endgroup$
$begingroup$
Thanks for your answer! This was exactly what I was looking for when I first thought of the question. I can't wait to study this function more.
$endgroup$
– Suchetan Dontha
Jan 8 at 0:34
$begingroup$
See my addition at the end of my answer.
$endgroup$
– marty cohen
Jan 8 at 0:43
add a comment |
$begingroup$
$I(a,b,c)
=sum_{n=0}^infty frac{1}{(an+b)^c}
$
Some simple cases.
$I(1,1,c)
=sum_{n=0}^infty frac{1}{(n+1)^c}
=sum_{n=1}^infty frac{1}{n^c}
=zeta(c)
$
$I(2,2,c)
=sum_{n=0}^infty frac{1}{(2n+2)^c}
=2^{-c}sum_{n=1}^infty frac{1}{n^c}
=2^{-c}zeta(c)
$
$begin{array}\
I(2,1,c)
&=sum_{n=0}^infty frac{1}{(2n+1)^c}\
&=sum_{n=0}^infty frac{1}{(2n+1)^c}+sum_{n=0}^infty frac{1}{(2n+2)^c}-sum_{n=0}^infty frac{1}{(2n+2)^c}\
&=sum_{n=1}^infty frac{1}{n^c}-I(2, 2, c)\
&=I(1, 1, c)-I(2, 2, c)\
&=zeta(c)-2^{-c}zeta(c)\
&=(1-2^{-c})zeta(c)\
end{array}
$
If $b$ is an integer,
$I(1,b,c)
=sum_{n=0}^infty frac{1}{(n+b)^c}
=sum_{n=b}^infty frac{1}{n^c}
=sum_{n=1}^infty frac{1}{n^c}-sum_{n=1}^{b-1} frac{1}{n^c}
=zeta(c)-sum_{n=1}^{b-1} frac{1}{n^c}
$
Note that
$I(a,b,c)
=sum_{n=0}^infty frac{1}{(an+b)^c}
=a^{-c}sum_{n=0}^infty frac{1}{(n+b/a)^c}
=a^{-c}zeta(c, b/a)
$
the Hurwitz zeta function.
See https://en.wikipedia.org/wiki/Hurwitz_zeta_function
for a discussion of this.
answered Jan 8 at 0:29
marty cohenmarty cohen
73.3k549128
$endgroup$
$begingroup$
Thanks for your answer! This was exactly what I was looking for when I first thought of the question. I can't wait to study this function more.
$endgroup$
– Suchetan Dontha
Jan 8 at 0:34
$begingroup$
See my addition at the end of my answer.
$endgroup$
– marty cohen
Jan 8 at 0:43
add a comment |
$begingroup$
$I(a,b,c)
=sum_{n=0}^infty frac{1}{(an+b)^c}
$
Some simple cases.
$I(1,1,c)
=sum_{n=0}^infty frac{1}{(n+1)^c}
=sum_{n=1}^infty frac{1}{n^c}
=zeta(c)
$
$I(2,2,c)
=sum_{n=0}^infty frac{1}{(2n+2)^c}
=2^{-c}sum_{n=1}^infty frac{1}{n^c}
=2^{-c}zeta(c)
$
$begin{array}\
I(2,1,c)
&=sum_{n=0}^infty frac{1}{(2n+1)^c}\
&=sum_{n=0}^infty frac{1}{(2n+1)^c}+sum_{n=0}^infty frac{1}{(2n+2)^c}-sum_{n=0}^infty frac{1}{(2n+2)^c}\
&=sum_{n=1}^infty frac{1}{n^c}-I(2, 2, c)\
&=I(1, 1, c)-I(2, 2, c)\
&=zeta(c)-2^{-c}zeta(c)\
&=(1-2^{-c})zeta(c)\
end{array}
$
If $b$ is an integer,
$I(1,b,c)
=sum_{n=0}^infty frac{1}{(n+b)^c}
=sum_{n=b}^infty frac{1}{n^c}
=sum_{n=1}^infty frac{1}{n^c}-sum_{n=1}^{b-1} frac{1}{n^c}
=zeta(c)-sum_{n=1}^{b-1} frac{1}{n^c}
$
Note that
$I(a,b,c)
=sum_{n=0}^infty frac{1}{(an+b)^c}
=a^{-c}sum_{n=0}^infty frac{1}{(n+b/a)^c}
=a^{-c}zeta(c, b/a)
$
the Hurwitz zeta function.
See https://en.wikipedia.org/wiki/Hurwitz_zeta_function
for a discussion of this.
answered Jan 8 at 0:29
marty cohenmarty cohen
73.3k549128
$endgroup$
$I(a,b,c)
=sum_{n=0}^infty frac{1}{(an+b)^c}
$
Some simple cases.
$I(1,1,c)
=sum_{n=0}^infty frac{1}{(n+1)^c}
=sum_{n=1}^infty frac{1}{n^c}
=zeta(c)
$
$I(2,2,c)
=sum_{n=0}^infty frac{1}{(2n+2)^c}
=2^{-c}sum_{n=1}^infty frac{1}{n^c}
=2^{-c}zeta(c)
$
$begin{array}\
I(2,1,c)
&=sum_{n=0}^infty frac{1}{(2n+1)^c}\
&=sum_{n=0}^infty frac{1}{(2n+1)^c}+sum_{n=0}^infty frac{1}{(2n+2)^c}-sum_{n=0}^infty frac{1}{(2n+2)^c}\
&=sum_{n=1}^infty frac{1}{n^c}-I(2, 2, c)\
&=I(1, 1, c)-I(2, 2, c)\
&=zeta(c)-2^{-c}zeta(c)\
&=(1-2^{-c})zeta(c)\
end{array}
$
If $b$ is an integer,
$I(1,b,c)
=sum_{n=0}^infty frac{1}{(n+b)^c}
=sum_{n=b}^infty frac{1}{n^c}
=sum_{n=1}^infty frac{1}{n^c}-sum_{n=1}^{b-1} frac{1}{n^c}
=zeta(c)-sum_{n=1}^{b-1} frac{1}{n^c}
$
Note that
$I(a,b,c)
=sum_{n=0}^infty frac{1}{(an+b)^c}
=a^{-c}sum_{n=0}^infty frac{1}{(n+b/a)^c}
=a^{-c}zeta(c, b/a)
$
the Hurwitz zeta function.
See https://en.wikipedia.org/wiki/Hurwitz_zeta_function
for a discussion of this.
answered Jan 8 at 0:29
marty cohenmarty cohen
73.3k549128
edited Jan 8 at 0:43
answered Jan 8 at 0:29
marty cohenmarty cohen
73.3k549128
answered Jan 8 at 0:29
marty cohenmarty cohen
73.3k549128
answered Jan 8 at 0:29
marty cohenmarty cohen
73.3k549128
73.3k549128
$begingroup$
Thanks for your answer! This was exactly what I was looking for when I first thought of the question. I can't wait to study this function more.
$endgroup$
– Suchetan Dontha
Jan 8 at 0:34
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See my addition at the end of my answer.
$endgroup$
– marty cohen
Jan 8 at 0:43
add a comment |
$begingroup$
Thanks for your answer! This was exactly what I was looking for when I first thought of the question. I can't wait to study this function more.
$endgroup$
– Suchetan Dontha
Jan 8 at 0:34
$begingroup$
See my addition at the end of my answer.
$endgroup$
– marty cohen
Jan 8 at 0:43
$begingroup$
Thanks for your answer! This was exactly what I was looking for when I first thought of the question. I can't wait to study this function more.
$endgroup$
– Suchetan Dontha
Jan 8 at 0:34
$begingroup$
Thanks for your answer! This was exactly what I was looking for when I first thought of the question. I can't wait to study this function more.
$endgroup$
– Suchetan Dontha
Jan 8 at 0:34
$begingroup$
See my addition at the end of my answer.
$endgroup$
– marty cohen
Jan 8 at 0:43
$begingroup$
See my addition at the end of my answer.
$endgroup$
– marty cohen
Jan 8 at 0:43
add a comment |
$begingroup$
While not a complete answer, if $c$ is a positive integer greater than 1 the sum can be written in terms of the polygamma function $psi^{(m)}(z)$.
From the series representation for the polygamma function, namely
$$psi^{(m)}(z) = (-1)^{m + 1} Gamma (m + 1) sum_{n = 0}^infty frac{1}{(n + z)^{m + 1}}, quad m > 0, z notin 0,mathbb{Z}^-$$
if we rewrite your sum as
$$I(a,b,c) = frac{1}{a^c} sum_{n = 0}^infty frac{1}{(n + b/a)^{(c - 1) + 1}},$$
in terms of the polygamma function the sum becomes
$$I(a,b,c) = frac{(-1)^c}{a^c Gamma (c)} psi^{(c - 1)} left (frac{b}{a} right ).$$
Here $c > 1$ where $c in mathbb{N}$, $a > 0$, and $b neq 0,-a,-2a,ldots$
answered Jan 8 at 1:36
omegadotomegadot
5,4422728
$endgroup$
add a comment |
$begingroup$
While not a complete answer, if $c$ is a positive integer greater than 1 the sum can be written in terms of the polygamma function $psi^{(m)}(z)$.
From the series representation for the polygamma function, namely
$$psi^{(m)}(z) = (-1)^{m + 1} Gamma (m + 1) sum_{n = 0}^infty frac{1}{(n + z)^{m + 1}}, quad m > 0, z notin 0,mathbb{Z}^-$$
if we rewrite your sum as
$$I(a,b,c) = frac{1}{a^c} sum_{n = 0}^infty frac{1}{(n + b/a)^{(c - 1) + 1}},$$
in terms of the polygamma function the sum becomes
$$I(a,b,c) = frac{(-1)^c}{a^c Gamma (c)} psi^{(c - 1)} left (frac{b}{a} right ).$$
Here $c > 1$ where $c in mathbb{N}$, $a > 0$, and $b neq 0,-a,-2a,ldots$
answered Jan 8 at 1:36
omegadotomegadot
5,4422728
$endgroup$
add a comment |
$begingroup$
While not a complete answer, if $c$ is a positive integer greater than 1 the sum can be written in terms of the polygamma function $psi^{(m)}(z)$.
From the series representation for the polygamma function, namely
$$psi^{(m)}(z) = (-1)^{m + 1} Gamma (m + 1) sum_{n = 0}^infty frac{1}{(n + z)^{m + 1}}, quad m > 0, z notin 0,mathbb{Z}^-$$
if we rewrite your sum as
$$I(a,b,c) = frac{1}{a^c} sum_{n = 0}^infty frac{1}{(n + b/a)^{(c - 1) + 1}},$$
in terms of the polygamma function the sum becomes
$$I(a,b,c) = frac{(-1)^c}{a^c Gamma (c)} psi^{(c - 1)} left (frac{b}{a} right ).$$
Here $c > 1$ where $c in mathbb{N}$, $a > 0$, and $b neq 0,-a,-2a,ldots$
answered Jan 8 at 1:36
omegadotomegadot
5,4422728
$endgroup$
While not a complete answer, if $c$ is a positive integer greater than 1 the sum can be written in terms of the polygamma function $psi^{(m)}(z)$.
From the series representation for the polygamma function, namely
$$psi^{(m)}(z) = (-1)^{m + 1} Gamma (m + 1) sum_{n = 0}^infty frac{1}{(n + z)^{m + 1}}, quad m > 0, z notin 0,mathbb{Z}^-$$
if we rewrite your sum as
$$I(a,b,c) = frac{1}{a^c} sum_{n = 0}^infty frac{1}{(n + b/a)^{(c - 1) + 1}},$$
in terms of the polygamma function the sum becomes
$$I(a,b,c) = frac{(-1)^c}{a^c Gamma (c)} psi^{(c - 1)} left (frac{b}{a} right ).$$
Here $c > 1$ where $c in mathbb{N}$, $a > 0$, and $b neq 0,-a,-2a,ldots$
answered Jan 8 at 1:36
omegadotomegadot
5,4422728
answered Jan 8 at 1:36
omegadotomegadot
5,4422728
answered Jan 8 at 1:36
omegadotomegadot
5,4422728
answered Jan 8 at 1:36
omegadotomegadot
5,4422728
5,4422728
add a comment |
add a comment |
$begingroup$
For $a=b=c=k$, where k is a constant, it seems as though $I(k, k, k) = frac{zeta(k)}{k^k}$
answered Jan 8 at 0:54
Suchetan DonthaSuchetan Dontha
14312
$endgroup$
add a comment |
$begingroup$
For $a=b=c=k$, where k is a constant, it seems as though $I(k, k, k) = frac{zeta(k)}{k^k}$
answered Jan 8 at 0:54
Suchetan DonthaSuchetan Dontha
14312
$endgroup$
add a comment |
$begingroup$
For $a=b=c=k$, where k is a constant, it seems as though $I(k, k, k) = frac{zeta(k)}{k^k}$
answered Jan 8 at 0:54
Suchetan DonthaSuchetan Dontha
14312
$endgroup$
For $a=b=c=k$, where k is a constant, it seems as though $I(k, k, k) = frac{zeta(k)}{k^k}$
answered Jan 8 at 0:54
Suchetan DonthaSuchetan Dontha
14312
answered Jan 8 at 0:54
Suchetan DonthaSuchetan Dontha
14312
answered Jan 8 at 0:54
Suchetan DonthaSuchetan Dontha
14312
answered Jan 8 at 0:54
Suchetan DonthaSuchetan Dontha
14312
14312
add a comment |
add a comment |
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$begingroup$
no, there isnt a closed formula for that
$endgroup$
– Masacroso
Jan 8 at 0:17
$begingroup$
Some special values can be obtained through the zeta function, e.g. $a=1, b=0$. For example, for $n in Bbb N$ $$zeta (2n)={frac {(-1)^{n+1}B_{2n}(2pi )^{2n}}{2(2n)!}}$$ $${displaystyle zeta (-n)=(-1)^{n}{frac {B_{n+1}}{n+1}}}$$ Other identities can be found at en.wikipedia.org/wiki/Riemann_zeta_function. .
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– Eevee Trainer
Jan 8 at 0:20
$begingroup$
But I imagine you already know the previous going by the tags and some of the values. To my knowledge there is no closed form known for odd integer inputs of the zeta function (maybe part of why Apery's constant, $zeta(3)$, is not known to be transcendental or not). Which, in light of that, (a) might mean this question will likely not yield anything more than partial answers and (b) might be better off on Math Overflow, but that's just a guess. At the very least it would be a very nontrivial question and I'm not the one to ask if things are better on MO.
$endgroup$
– Eevee Trainer
Jan 8 at 0:20
$begingroup$
@EeveeTrainer Thank you for your comment. I've edited the question to be more open and reasonable. Further, I'm not too knowledgable about the zeta function given my high school calculus knowledge. Any information would help me!
$endgroup$
– Suchetan Dontha
Jan 8 at 0:22
$begingroup$
With respect to one particular case ($a=b=1, c=2$), there is a discussion on it in this video - youtube.com/watch?v=aYc9ManKC-s - I haven't watched it yet so I don't have anything to offer there. Just figured you might find it useful.
$endgroup$
– Eevee Trainer
Jan 8 at 0:25