What is this sum equal to? $sigma(n)=sum_ineq j frac1i^n j^n$
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4
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I have recently come across the following sum, taken over all positive integers $i$ and $j$ such that $i neq j$:
$$
sigma(n)=sum_ineq j frac1i^n j^n,
$$
where $n$ is a positive integer greater than $1$.
Can this somehow be written in terms of the Riemann Zeta function? Is there already a zeta function of this kind?
As a corollary question, what about the sum
$$
sigma(n)=sum_ineq j frac1i^n j^n-1?
$$
sequences-and-series convergence riemann-zeta
edited Nov 8 at 13:43
Daniel R
2,45732035
asked Nov 8 at 7:31
Flermat
1,19211129
add a comment |
up vote
4
down vote
favorite
I have recently come across the following sum, taken over all positive integers $i$ and $j$ such that $i neq j$:
$$
sigma(n)=sum_ineq j frac1i^n j^n,
$$
where $n$ is a positive integer greater than $1$.
Can this somehow be written in terms of the Riemann Zeta function? Is there already a zeta function of this kind?
As a corollary question, what about the sum
$$
sigma(n)=sum_ineq j frac1i^n j^n-1?
$$
sequences-and-series convergence riemann-zeta
edited Nov 8 at 13:43
Daniel R
2,45732035
asked Nov 8 at 7:31
Flermat
1,19211129
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I have recently come across the following sum, taken over all positive integers $i$ and $j$ such that $i neq j$:
$$
sigma(n)=sum_ineq j frac1i^n j^n,
$$
where $n$ is a positive integer greater than $1$.
Can this somehow be written in terms of the Riemann Zeta function? Is there already a zeta function of this kind?
As a corollary question, what about the sum
$$
sigma(n)=sum_ineq j frac1i^n j^n-1?
$$
sequences-and-series convergence riemann-zeta
edited Nov 8 at 13:43
Daniel R
2,45732035
asked Nov 8 at 7:31
Flermat
1,19211129
I have recently come across the following sum, taken over all positive integers $i$ and $j$ such that $i neq j$:
$$
sigma(n)=sum_ineq j frac1i^n j^n,
$$
where $n$ is a positive integer greater than $1$.
Can this somehow be written in terms of the Riemann Zeta function? Is there already a zeta function of this kind?
As a corollary question, what about the sum
$$
sigma(n)=sum_ineq j frac1i^n j^n-1?
$$
sequences-and-series convergence riemann-zeta
sequences-and-series convergence riemann-zeta
edited Nov 8 at 13:43
Daniel R
2,45732035
asked Nov 8 at 7:31
Flermat
1,19211129
edited Nov 8 at 13:43
Daniel R
2,45732035
asked Nov 8 at 7:31
Flermat
1,19211129
edited Nov 8 at 13:43
Daniel R
2,45732035
edited Nov 8 at 13:43
Daniel R
2,45732035
edited Nov 8 at 13:43
Daniel R
2,45732035
2,45732035
asked Nov 8 at 7:31
Flermat
1,19211129
asked Nov 8 at 7:31
Flermat
1,19211129
asked Nov 8 at 7:31
Flermat
1,19211129
1,19211129
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
8
down vote
accepted
Your first sum equals
$$
sum_i,jfrac1i^nj^n-sum_i=jfrac1i^nj^n=zeta(n)^2-zeta(2n).$$
The same trick works for your the second sum.
answered Nov 8 at 7:32
Lord Shark the Unknown
96.1k957125
Thank you for your response. What about the second case, i.e., when the $i$ and $j$ in the denominator have differing exponents?
– Flermat
Nov 8 at 7:44
1
As I say, the same trick works for your the second sum. @Flermat
– Lord Shark the Unknown
Nov 8 at 7:48
Oh, my bad, didn't notice that. Thank you.
– Flermat
Nov 8 at 7:52
add a comment |
up vote
3
down vote
HINT
We have that
$$sigma(n)=sum_ineq j frac1i^n j^n=sum_i frac1i^nsum_j frac1 j^n-sum_k frac1k^2n$$
answered Nov 8 at 7:34
gimusi
83.5k74292
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Your first sum equals
$$
sum_i,jfrac1i^nj^n-sum_i=jfrac1i^nj^n=zeta(n)^2-zeta(2n).$$
The same trick works for your the second sum.
answered Nov 8 at 7:32
Lord Shark the Unknown
96.1k957125
Thank you for your response. What about the second case, i.e., when the $i$ and $j$ in the denominator have differing exponents?
– Flermat
Nov 8 at 7:44
1
As I say, the same trick works for your the second sum. @Flermat
– Lord Shark the Unknown
Nov 8 at 7:48
Oh, my bad, didn't notice that. Thank you.
– Flermat
Nov 8 at 7:52
add a comment |
up vote
8
down vote
accepted
Your first sum equals
$$
sum_i,jfrac1i^nj^n-sum_i=jfrac1i^nj^n=zeta(n)^2-zeta(2n).$$
The same trick works for your the second sum.
answered Nov 8 at 7:32
Lord Shark the Unknown
96.1k957125
Thank you for your response. What about the second case, i.e., when the $i$ and $j$ in the denominator have differing exponents?
– Flermat
Nov 8 at 7:44
1
As I say, the same trick works for your the second sum. @Flermat
– Lord Shark the Unknown
Nov 8 at 7:48
Oh, my bad, didn't notice that. Thank you.
– Flermat
Nov 8 at 7:52
add a comment |
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Your first sum equals
$$
sum_i,jfrac1i^nj^n-sum_i=jfrac1i^nj^n=zeta(n)^2-zeta(2n).$$
The same trick works for your the second sum.
answered Nov 8 at 7:32
Lord Shark the Unknown
96.1k957125
Your first sum equals
$$
sum_i,jfrac1i^nj^n-sum_i=jfrac1i^nj^n=zeta(n)^2-zeta(2n).$$
The same trick works for your the second sum.
answered Nov 8 at 7:32
Lord Shark the Unknown
96.1k957125
edited Nov 8 at 7:45
answered Nov 8 at 7:32
Lord Shark the Unknown
96.1k957125
answered Nov 8 at 7:32
Lord Shark the Unknown
96.1k957125
answered Nov 8 at 7:32
Lord Shark the Unknown
96.1k957125
96.1k957125
Thank you for your response. What about the second case, i.e., when the $i$ and $j$ in the denominator have differing exponents?
– Flermat
Nov 8 at 7:44
1
As I say, the same trick works for your the second sum. @Flermat
– Lord Shark the Unknown
Nov 8 at 7:48
Oh, my bad, didn't notice that. Thank you.
– Flermat
Nov 8 at 7:52
add a comment |
Thank you for your response. What about the second case, i.e., when the $i$ and $j$ in the denominator have differing exponents?
– Flermat
Nov 8 at 7:44
1
As I say, the same trick works for your the second sum. @Flermat
– Lord Shark the Unknown
Nov 8 at 7:48
Oh, my bad, didn't notice that. Thank you.
– Flermat
Nov 8 at 7:52
Thank you for your response. What about the second case, i.e., when the $i$ and $j$ in the denominator have differing exponents?
– Flermat
Nov 8 at 7:44
Thank you for your response. What about the second case, i.e., when the $i$ and $j$ in the denominator have differing exponents?
– Flermat
Nov 8 at 7:44
1
1
As I say, the same trick works for your the second sum. @Flermat
– Lord Shark the Unknown
Nov 8 at 7:48
As I say, the same trick works for your the second sum. @Flermat
– Lord Shark the Unknown
Nov 8 at 7:48
Oh, my bad, didn't notice that. Thank you.
– Flermat
Nov 8 at 7:52
Oh, my bad, didn't notice that. Thank you.
– Flermat
Nov 8 at 7:52
add a comment |
up vote
3
down vote
HINT
We have that
$$sigma(n)=sum_ineq j frac1i^n j^n=sum_i frac1i^nsum_j frac1 j^n-sum_k frac1k^2n$$
answered Nov 8 at 7:34
gimusi
83.5k74292
add a comment |
up vote
3
down vote
HINT
We have that
$$sigma(n)=sum_ineq j frac1i^n j^n=sum_i frac1i^nsum_j frac1 j^n-sum_k frac1k^2n$$
answered Nov 8 at 7:34
gimusi
83.5k74292
add a comment |
up vote
3
down vote
up vote
3
down vote
HINT
We have that
$$sigma(n)=sum_ineq j frac1i^n j^n=sum_i frac1i^nsum_j frac1 j^n-sum_k frac1k^2n$$
answered Nov 8 at 7:34
gimusi
83.5k74292
HINT
We have that
$$sigma(n)=sum_ineq j frac1i^n j^n=sum_i frac1i^nsum_j frac1 j^n-sum_k frac1k^2n$$
answered Nov 8 at 7:34
gimusi
83.5k74292
edited Nov 8 at 7:36
answered Nov 8 at 7:34
gimusi
83.5k74292
answered Nov 8 at 7:34
gimusi
83.5k74292
answered Nov 8 at 7:34
gimusi
83.5k74292
83.5k74292
add a comment |
add a comment |
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