Mixing
Find whether : $sum_{ I subset mathbb{N}} e^{-sqrt{S(I)}}$ converges
Does the following sum converges : $$sum_{ I subset mathbb{N}} e^{-sqrt{S(I)}}$$
where : $$ S(I) =sum_{ i in I} i$$
I don’t know how to approach this problem. Nevertheless, maybe it’s possible to have some intuition about the problem.
If $mid I mid = n$ then we know that the $S(I)$ which are going to give an important weight to the sum are in $O(n^2)$. Hence maybe luckily :
$$sum_{I subset mathbb{N}, S(I) = O(mid I mid ^2)} e^{-sqrt{S(I)}}$$
has the same nature of our sum.
In this case the square root makes it easy and the sum converges.
Yet it’s possible that this intuition is false since there are a lot of $S(I) ne O(mid I mid ^2)$ (infinitely many actually) so it might give a lot of weight to the sum... I don’t really know.
Note that in order that $S(I)$ makes sens, we have : $I < infty$.
calculus real-analysis integration sequences-and-series limits
asked Nov 21 '18 at 17:45
Interesting problemsInteresting problems
13310
add a comment |
Does the following sum converges : $$sum_{ I subset mathbb{N}} e^{-sqrt{S(I)}}$$
where : $$ S(I) =sum_{ i in I} i$$
I don’t know how to approach this problem. Nevertheless, maybe it’s possible to have some intuition about the problem.
If $mid I mid = n$ then we know that the $S(I)$ which are going to give an important weight to the sum are in $O(n^2)$. Hence maybe luckily :
$$sum_{I subset mathbb{N}, S(I) = O(mid I mid ^2)} e^{-sqrt{S(I)}}$$
has the same nature of our sum.
In this case the square root makes it easy and the sum converges.
Yet it’s possible that this intuition is false since there are a lot of $S(I) ne O(mid I mid ^2)$ (infinitely many actually) so it might give a lot of weight to the sum... I don’t really know.
Note that in order that $S(I)$ makes sens, we have : $I < infty$.
calculus real-analysis integration sequences-and-series limits
asked Nov 21 '18 at 17:45
Interesting problemsInteresting problems
13310
Can you confirm that $mid I mid$ is in $mathbb{R}$, I think it’s since otherwise $S(I)$ doesn’t make sens, but it’s just to be sure.
– Thinking
Nov 21 '18 at 18:22
Yes, $I < infty$.
– Interesting problems
Nov 21 '18 at 18:49
add a comment |
Does the following sum converges : $$sum_{ I subset mathbb{N}} e^{-sqrt{S(I)}}$$
where : $$ S(I) =sum_{ i in I} i$$
I don’t know how to approach this problem. Nevertheless, maybe it’s possible to have some intuition about the problem.
If $mid I mid = n$ then we know that the $S(I)$ which are going to give an important weight to the sum are in $O(n^2)$. Hence maybe luckily :
$$sum_{I subset mathbb{N}, S(I) = O(mid I mid ^2)} e^{-sqrt{S(I)}}$$
has the same nature of our sum.
In this case the square root makes it easy and the sum converges.
Yet it’s possible that this intuition is false since there are a lot of $S(I) ne O(mid I mid ^2)$ (infinitely many actually) so it might give a lot of weight to the sum... I don’t really know.
Note that in order that $S(I)$ makes sens, we have : $I < infty$.
calculus real-analysis integration sequences-and-series limits
asked Nov 21 '18 at 17:45
Interesting problemsInteresting problems
13310
Does the following sum converges : $$sum_{ I subset mathbb{N}} e^{-sqrt{S(I)}}$$
where : $$ S(I) =sum_{ i in I} i$$
I don’t know how to approach this problem. Nevertheless, maybe it’s possible to have some intuition about the problem.
If $mid I mid = n$ then we know that the $S(I)$ which are going to give an important weight to the sum are in $O(n^2)$. Hence maybe luckily :
$$sum_{I subset mathbb{N}, S(I) = O(mid I mid ^2)} e^{-sqrt{S(I)}}$$
has the same nature of our sum.
In this case the square root makes it easy and the sum converges.
Yet it’s possible that this intuition is false since there are a lot of $S(I) ne O(mid I mid ^2)$ (infinitely many actually) so it might give a lot of weight to the sum... I don’t really know.
Note that in order that $S(I)$ makes sens, we have : $I < infty$.
calculus real-analysis integration sequences-and-series limits
calculus real-analysis integration sequences-and-series limits
asked Nov 21 '18 at 17:45
Interesting problemsInteresting problems
13310
asked Nov 21 '18 at 17:45
Interesting problemsInteresting problems
13310
edited Nov 21 '18 at 18:51
Interesting problems
asked Nov 21 '18 at 17:45
Interesting problemsInteresting problems
13310
asked Nov 21 '18 at 17:45
Interesting problemsInteresting problems
13310
asked Nov 21 '18 at 17:45
Interesting problemsInteresting problems
13310
13310
Can you confirm that $mid I mid$ is in $mathbb{R}$, I think it’s since otherwise $S(I)$ doesn’t make sens, but it’s just to be sure.
– Thinking
Nov 21 '18 at 18:22
Yes, $I < infty$.
– Interesting problems
Nov 21 '18 at 18:49
add a comment |
Can you confirm that $mid I mid$ is in $mathbb{R}$, I think it’s since otherwise $S(I)$ doesn’t make sens, but it’s just to be sure.
– Thinking
Nov 21 '18 at 18:22
Yes, $I < infty$.
– Interesting problems
Nov 21 '18 at 18:49
Can you confirm that $mid I mid$ is in $mathbb{R}$, I think it’s since otherwise $S(I)$ doesn’t make sens, but it’s just to be sure.
– Thinking
Nov 21 '18 at 18:22
Can you confirm that $mid I mid$ is in $mathbb{R}$, I think it’s since otherwise $S(I)$ doesn’t make sens, but it’s just to be sure.
– Thinking
Nov 21 '18 at 18:22
Yes, $I < infty$.
– Interesting problems
Nov 21 '18 at 18:49
Yes, $I < infty$.
– Interesting problems
Nov 21 '18 at 18:49
add a comment |
2 Answers
2
active
oldest
votes
To make things precise, when $|I| = infty$, we redefine the value of $S(I)$ as $+infty$ and $e^{-sqrt{S(I)}} = 0$.
After this change, the summand inside the sum $sumlimits_{Isubset mathbb{N}} e^{-sqrt{S(I)}}$ will be non-zero for countably many $I$. Since all summands are non-negative, the sum is well defined and takes value in $[0,infty]$. Furthermore, we can compute it by enumerating those $I$ with $|I| < infty$ in arbitrary order and get the same result. As a result,
$$begin{align}sum_{Isubset mathbb{N}} e^{-sqrt{S(I)}}
stackrel{def}{=} sum_{Isubset mathbb{N}, |I| < infty} e^{-sqrt{S(I)}}
&= 2sum_{Isubset mathbb{Z}_{+}, |I| < infty} e^{-sqrt{S(I)}}
= 2sum_{n=0}^infty sum_{I subset mathbb{Z}_{+}, S(I) = n} e^{-sqrt{n}}\
&= 2sum_{n=0}^infty q(n) e^{-sqrt{n}}
end{align}
$$
where $q(n) = | { I subset mathbb{Z}_{+} : S(I) = n } |$ is the number of partitions
of integer $n$ into distinct parts.
The OGF of $q(n)$ equals to
$$sum_{n=0}^infty q(n) z^n = prod_{k=1}^infty ( 1 + z^k )$$
The closed form of $q(n)$ is not known. However, we do know for large $n$,${}^{color{blue}{[1]}}$
$$q(n) sim frac{3^{3/4}}{12 n^{3/4}} expleft(pisqrt{frac{n}{3}}right) $$
Since $alpha stackrel{def}{=} frac{pi}{sqrt{3}} - 1 > 0$, the sub-sum over those $I$ with $S(I) = n$ blows up like $e^{alphasqrt{n}}$. From this, we can deduce
$$sum_{Isubset mathbb{N}} e^{-sqrt{S(I)}} = infty$$
Refs
$color{blue}{[1]}$ -
Philippe Flajolet, Robert Sedgewick
Analytic Combinatorics,
Cambridge University Press; (1st ed., 2009). Formula found at VIII.6 Saddle-point asymptotics / Integer partitions.
answered Nov 21 '18 at 19:11
achille huiachille hui
95.6k5130257
1
It's quite nice to see that when the exponent is $ > 1/2$ then the serie converges. (+1)
– Thinking
Nov 21 '18 at 19:21
What do you mean "equivalent of $q(n)$"? If $S(I) = n$, then those $i in I$ gives a partition of $n$ into distinct parts and vise versa. So the number of those $I$ with $S(I) = n$ is precisely the number of partitions of $n$ into distinct parts.
– achille hui
Nov 21 '18 at 19:26
1
I was talking about the result $ q(n) sim ...$, but it's ok I found an article on wiki about this asymptotic
– Thinking
Nov 21 '18 at 19:28
@reuns ????? it is $e^{pi sqrt{frac{color{red}{n}}{3}}}$
– achille hui
Nov 21 '18 at 19:46
On wiki it says $q(n) sim O(n^l) c^{sqrt{n}}$ with $c = e^{pi sqrt{2/3}}$ and the above series diverges because $c > e$. The trivial argument in my deleted answer shows that $c ge 2^{sqrt{2}}$ that's why I ask where $e^{pi sqrt{2/3}}$ comes from.
– reuns
Nov 21 '18 at 19:53
|
show 2 more comments
That sum, or just any sum with more than countably many positive summands cannot converge:
Let $X$ be a set and $fcolon XtoBbb R$ a map such that $f(x)> 0$ fore all $xin X$. Assume that $sum_{xin X}f(x)$ converges, say $sum_{xin X}f(x)=Sin Bbb R$. Then for $Ysubseteq X$, clearly $sum_{xin Y}f(x)le sum _{xin X}f(x)$.
For $nin Bbb N$, let $X_n={,xin Xmid f(x)>frac 1n,}$. Then $$Mge sum_{xin X_n}f(x)ge sum_{xin X_n}frac 1n=frac 1n|X_n|$$
and in particular, $X_n$ must be finite. As $X=bigcup_{ninBbb N}X_n$, we conclude that $X$ is countable.
answered Nov 21 '18 at 18:05
Hagen von EitzenHagen von Eitzen
276k21269496
I think that the OP was talking about $I$ finite, in this case this is countable. Otherwise $S(I)$ doesn’t make sens.
– Thinking
Nov 21 '18 at 18:20
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
To make things precise, when $|I| = infty$, we redefine the value of $S(I)$ as $+infty$ and $e^{-sqrt{S(I)}} = 0$.
After this change, the summand inside the sum $sumlimits_{Isubset mathbb{N}} e^{-sqrt{S(I)}}$ will be non-zero for countably many $I$. Since all summands are non-negative, the sum is well defined and takes value in $[0,infty]$. Furthermore, we can compute it by enumerating those $I$ with $|I| < infty$ in arbitrary order and get the same result. As a result,
$$begin{align}sum_{Isubset mathbb{N}} e^{-sqrt{S(I)}}
stackrel{def}{=} sum_{Isubset mathbb{N}, |I| < infty} e^{-sqrt{S(I)}}
&= 2sum_{Isubset mathbb{Z}_{+}, |I| < infty} e^{-sqrt{S(I)}}
= 2sum_{n=0}^infty sum_{I subset mathbb{Z}_{+}, S(I) = n} e^{-sqrt{n}}\
&= 2sum_{n=0}^infty q(n) e^{-sqrt{n}}
end{align}
$$
where $q(n) = | { I subset mathbb{Z}_{+} : S(I) = n } |$ is the number of partitions
of integer $n$ into distinct parts.
The OGF of $q(n)$ equals to
$$sum_{n=0}^infty q(n) z^n = prod_{k=1}^infty ( 1 + z^k )$$
The closed form of $q(n)$ is not known. However, we do know for large $n$,${}^{color{blue}{[1]}}$
$$q(n) sim frac{3^{3/4}}{12 n^{3/4}} expleft(pisqrt{frac{n}{3}}right) $$
Since $alpha stackrel{def}{=} frac{pi}{sqrt{3}} - 1 > 0$, the sub-sum over those $I$ with $S(I) = n$ blows up like $e^{alphasqrt{n}}$. From this, we can deduce
$$sum_{Isubset mathbb{N}} e^{-sqrt{S(I)}} = infty$$
Refs
$color{blue}{[1]}$ -
Philippe Flajolet, Robert Sedgewick
Analytic Combinatorics,
Cambridge University Press; (1st ed., 2009). Formula found at VIII.6 Saddle-point asymptotics / Integer partitions.
answered Nov 21 '18 at 19:11
achille huiachille hui
95.6k5130257
1
It's quite nice to see that when the exponent is $ > 1/2$ then the serie converges. (+1)
– Thinking
Nov 21 '18 at 19:21
What do you mean "equivalent of $q(n)$"? If $S(I) = n$, then those $i in I$ gives a partition of $n$ into distinct parts and vise versa. So the number of those $I$ with $S(I) = n$ is precisely the number of partitions of $n$ into distinct parts.
– achille hui
Nov 21 '18 at 19:26
1
I was talking about the result $ q(n) sim ...$, but it's ok I found an article on wiki about this asymptotic
– Thinking
Nov 21 '18 at 19:28
@reuns ????? it is $e^{pi sqrt{frac{color{red}{n}}{3}}}$
– achille hui
Nov 21 '18 at 19:46
On wiki it says $q(n) sim O(n^l) c^{sqrt{n}}$ with $c = e^{pi sqrt{2/3}}$ and the above series diverges because $c > e$. The trivial argument in my deleted answer shows that $c ge 2^{sqrt{2}}$ that's why I ask where $e^{pi sqrt{2/3}}$ comes from.
– reuns
Nov 21 '18 at 19:53
|
show 2 more comments
To make things precise, when $|I| = infty$, we redefine the value of $S(I)$ as $+infty$ and $e^{-sqrt{S(I)}} = 0$.
After this change, the summand inside the sum $sumlimits_{Isubset mathbb{N}} e^{-sqrt{S(I)}}$ will be non-zero for countably many $I$. Since all summands are non-negative, the sum is well defined and takes value in $[0,infty]$. Furthermore, we can compute it by enumerating those $I$ with $|I| < infty$ in arbitrary order and get the same result. As a result,
$$begin{align}sum_{Isubset mathbb{N}} e^{-sqrt{S(I)}}
stackrel{def}{=} sum_{Isubset mathbb{N}, |I| < infty} e^{-sqrt{S(I)}}
&= 2sum_{Isubset mathbb{Z}_{+}, |I| < infty} e^{-sqrt{S(I)}}
= 2sum_{n=0}^infty sum_{I subset mathbb{Z}_{+}, S(I) = n} e^{-sqrt{n}}\
&= 2sum_{n=0}^infty q(n) e^{-sqrt{n}}
end{align}
$$
where $q(n) = | { I subset mathbb{Z}_{+} : S(I) = n } |$ is the number of partitions
of integer $n$ into distinct parts.
The OGF of $q(n)$ equals to
$$sum_{n=0}^infty q(n) z^n = prod_{k=1}^infty ( 1 + z^k )$$
The closed form of $q(n)$ is not known. However, we do know for large $n$,${}^{color{blue}{[1]}}$
$$q(n) sim frac{3^{3/4}}{12 n^{3/4}} expleft(pisqrt{frac{n}{3}}right) $$
Since $alpha stackrel{def}{=} frac{pi}{sqrt{3}} - 1 > 0$, the sub-sum over those $I$ with $S(I) = n$ blows up like $e^{alphasqrt{n}}$. From this, we can deduce
$$sum_{Isubset mathbb{N}} e^{-sqrt{S(I)}} = infty$$
Refs
$color{blue}{[1]}$ -
Philippe Flajolet, Robert Sedgewick
Analytic Combinatorics,
Cambridge University Press; (1st ed., 2009). Formula found at VIII.6 Saddle-point asymptotics / Integer partitions.
answered Nov 21 '18 at 19:11
achille huiachille hui
95.6k5130257
1
It's quite nice to see that when the exponent is $ > 1/2$ then the serie converges. (+1)
– Thinking
Nov 21 '18 at 19:21
What do you mean "equivalent of $q(n)$"? If $S(I) = n$, then those $i in I$ gives a partition of $n$ into distinct parts and vise versa. So the number of those $I$ with $S(I) = n$ is precisely the number of partitions of $n$ into distinct parts.
– achille hui
Nov 21 '18 at 19:26
1
I was talking about the result $ q(n) sim ...$, but it's ok I found an article on wiki about this asymptotic
– Thinking
Nov 21 '18 at 19:28
@reuns ????? it is $e^{pi sqrt{frac{color{red}{n}}{3}}}$
– achille hui
Nov 21 '18 at 19:46
On wiki it says $q(n) sim O(n^l) c^{sqrt{n}}$ with $c = e^{pi sqrt{2/3}}$ and the above series diverges because $c > e$. The trivial argument in my deleted answer shows that $c ge 2^{sqrt{2}}$ that's why I ask where $e^{pi sqrt{2/3}}$ comes from.
– reuns
Nov 21 '18 at 19:53
|
show 2 more comments
To make things precise, when $|I| = infty$, we redefine the value of $S(I)$ as $+infty$ and $e^{-sqrt{S(I)}} = 0$.
After this change, the summand inside the sum $sumlimits_{Isubset mathbb{N}} e^{-sqrt{S(I)}}$ will be non-zero for countably many $I$. Since all summands are non-negative, the sum is well defined and takes value in $[0,infty]$. Furthermore, we can compute it by enumerating those $I$ with $|I| < infty$ in arbitrary order and get the same result. As a result,
$$begin{align}sum_{Isubset mathbb{N}} e^{-sqrt{S(I)}}
stackrel{def}{=} sum_{Isubset mathbb{N}, |I| < infty} e^{-sqrt{S(I)}}
&= 2sum_{Isubset mathbb{Z}_{+}, |I| < infty} e^{-sqrt{S(I)}}
= 2sum_{n=0}^infty sum_{I subset mathbb{Z}_{+}, S(I) = n} e^{-sqrt{n}}\
&= 2sum_{n=0}^infty q(n) e^{-sqrt{n}}
end{align}
$$
where $q(n) = | { I subset mathbb{Z}_{+} : S(I) = n } |$ is the number of partitions
of integer $n$ into distinct parts.
The OGF of $q(n)$ equals to
$$sum_{n=0}^infty q(n) z^n = prod_{k=1}^infty ( 1 + z^k )$$
The closed form of $q(n)$ is not known. However, we do know for large $n$,${}^{color{blue}{[1]}}$
$$q(n) sim frac{3^{3/4}}{12 n^{3/4}} expleft(pisqrt{frac{n}{3}}right) $$
Since $alpha stackrel{def}{=} frac{pi}{sqrt{3}} - 1 > 0$, the sub-sum over those $I$ with $S(I) = n$ blows up like $e^{alphasqrt{n}}$. From this, we can deduce
$$sum_{Isubset mathbb{N}} e^{-sqrt{S(I)}} = infty$$
Refs
$color{blue}{[1]}$ -
Philippe Flajolet, Robert Sedgewick
Analytic Combinatorics,
Cambridge University Press; (1st ed., 2009). Formula found at VIII.6 Saddle-point asymptotics / Integer partitions.
answered Nov 21 '18 at 19:11
achille huiachille hui
95.6k5130257
To make things precise, when $|I| = infty$, we redefine the value of $S(I)$ as $+infty$ and $e^{-sqrt{S(I)}} = 0$.
After this change, the summand inside the sum $sumlimits_{Isubset mathbb{N}} e^{-sqrt{S(I)}}$ will be non-zero for countably many $I$. Since all summands are non-negative, the sum is well defined and takes value in $[0,infty]$. Furthermore, we can compute it by enumerating those $I$ with $|I| < infty$ in arbitrary order and get the same result. As a result,
$$begin{align}sum_{Isubset mathbb{N}} e^{-sqrt{S(I)}}
stackrel{def}{=} sum_{Isubset mathbb{N}, |I| < infty} e^{-sqrt{S(I)}}
&= 2sum_{Isubset mathbb{Z}_{+}, |I| < infty} e^{-sqrt{S(I)}}
= 2sum_{n=0}^infty sum_{I subset mathbb{Z}_{+}, S(I) = n} e^{-sqrt{n}}\
&= 2sum_{n=0}^infty q(n) e^{-sqrt{n}}
end{align}
$$
where $q(n) = | { I subset mathbb{Z}_{+} : S(I) = n } |$ is the number of partitions
of integer $n$ into distinct parts.
The OGF of $q(n)$ equals to
$$sum_{n=0}^infty q(n) z^n = prod_{k=1}^infty ( 1 + z^k )$$
The closed form of $q(n)$ is not known. However, we do know for large $n$,${}^{color{blue}{[1]}}$
$$q(n) sim frac{3^{3/4}}{12 n^{3/4}} expleft(pisqrt{frac{n}{3}}right) $$
Since $alpha stackrel{def}{=} frac{pi}{sqrt{3}} - 1 > 0$, the sub-sum over those $I$ with $S(I) = n$ blows up like $e^{alphasqrt{n}}$. From this, we can deduce
$$sum_{Isubset mathbb{N}} e^{-sqrt{S(I)}} = infty$$
Refs
$color{blue}{[1]}$ -
Philippe Flajolet, Robert Sedgewick
Analytic Combinatorics,
Cambridge University Press; (1st ed., 2009). Formula found at VIII.6 Saddle-point asymptotics / Integer partitions.
answered Nov 21 '18 at 19:11
achille huiachille hui
95.6k5130257
edited Nov 21 '18 at 19:44
answered Nov 21 '18 at 19:11
achille huiachille hui
95.6k5130257
answered Nov 21 '18 at 19:11
achille huiachille hui
95.6k5130257
answered Nov 21 '18 at 19:11
achille huiachille hui
95.6k5130257
95.6k5130257
1
It's quite nice to see that when the exponent is $ > 1/2$ then the serie converges. (+1)
– Thinking
Nov 21 '18 at 19:21
What do you mean "equivalent of $q(n)$"? If $S(I) = n$, then those $i in I$ gives a partition of $n$ into distinct parts and vise versa. So the number of those $I$ with $S(I) = n$ is precisely the number of partitions of $n$ into distinct parts.
– achille hui
Nov 21 '18 at 19:26
1
I was talking about the result $ q(n) sim ...$, but it's ok I found an article on wiki about this asymptotic
– Thinking
Nov 21 '18 at 19:28
@reuns ????? it is $e^{pi sqrt{frac{color{red}{n}}{3}}}$
– achille hui
Nov 21 '18 at 19:46
On wiki it says $q(n) sim O(n^l) c^{sqrt{n}}$ with $c = e^{pi sqrt{2/3}}$ and the above series diverges because $c > e$. The trivial argument in my deleted answer shows that $c ge 2^{sqrt{2}}$ that's why I ask where $e^{pi sqrt{2/3}}$ comes from.
– reuns
Nov 21 '18 at 19:53
|
show 2 more comments
1
It's quite nice to see that when the exponent is $ > 1/2$ then the serie converges. (+1)
– Thinking
Nov 21 '18 at 19:21
What do you mean "equivalent of $q(n)$"? If $S(I) = n$, then those $i in I$ gives a partition of $n$ into distinct parts and vise versa. So the number of those $I$ with $S(I) = n$ is precisely the number of partitions of $n$ into distinct parts.
– achille hui
Nov 21 '18 at 19:26
1
I was talking about the result $ q(n) sim ...$, but it's ok I found an article on wiki about this asymptotic
– Thinking
Nov 21 '18 at 19:28
@reuns ????? it is $e^{pi sqrt{frac{color{red}{n}}{3}}}$
– achille hui
Nov 21 '18 at 19:46
On wiki it says $q(n) sim O(n^l) c^{sqrt{n}}$ with $c = e^{pi sqrt{2/3}}$ and the above series diverges because $c > e$. The trivial argument in my deleted answer shows that $c ge 2^{sqrt{2}}$ that's why I ask where $e^{pi sqrt{2/3}}$ comes from.
– reuns
Nov 21 '18 at 19:53
1
1
It's quite nice to see that when the exponent is $ > 1/2$ then the serie converges. (+1)
– Thinking
Nov 21 '18 at 19:21
It's quite nice to see that when the exponent is $ > 1/2$ then the serie converges. (+1)
– Thinking
Nov 21 '18 at 19:21
What do you mean "equivalent of $q(n)$"? If $S(I) = n$, then those $i in I$ gives a partition of $n$ into distinct parts and vise versa. So the number of those $I$ with $S(I) = n$ is precisely the number of partitions of $n$ into distinct parts.
– achille hui
Nov 21 '18 at 19:26
What do you mean "equivalent of $q(n)$"? If $S(I) = n$, then those $i in I$ gives a partition of $n$ into distinct parts and vise versa. So the number of those $I$ with $S(I) = n$ is precisely the number of partitions of $n$ into distinct parts.
– achille hui
Nov 21 '18 at 19:26
1
1
I was talking about the result $ q(n) sim ...$, but it's ok I found an article on wiki about this asymptotic
– Thinking
Nov 21 '18 at 19:28
I was talking about the result $ q(n) sim ...$, but it's ok I found an article on wiki about this asymptotic
– Thinking
Nov 21 '18 at 19:28
@reuns ????? it is $e^{pi sqrt{frac{color{red}{n}}{3}}}$
– achille hui
Nov 21 '18 at 19:46
@reuns ????? it is $e^{pi sqrt{frac{color{red}{n}}{3}}}$
– achille hui
Nov 21 '18 at 19:46
On wiki it says $q(n) sim O(n^l) c^{sqrt{n}}$ with $c = e^{pi sqrt{2/3}}$ and the above series diverges because $c > e$. The trivial argument in my deleted answer shows that $c ge 2^{sqrt{2}}$ that's why I ask where $e^{pi sqrt{2/3}}$ comes from.
– reuns
Nov 21 '18 at 19:53
On wiki it says $q(n) sim O(n^l) c^{sqrt{n}}$ with $c = e^{pi sqrt{2/3}}$ and the above series diverges because $c > e$. The trivial argument in my deleted answer shows that $c ge 2^{sqrt{2}}$ that's why I ask where $e^{pi sqrt{2/3}}$ comes from.
– reuns
Nov 21 '18 at 19:53
|
show 2 more comments
That sum, or just any sum with more than countably many positive summands cannot converge:
Let $X$ be a set and $fcolon XtoBbb R$ a map such that $f(x)> 0$ fore all $xin X$. Assume that $sum_{xin X}f(x)$ converges, say $sum_{xin X}f(x)=Sin Bbb R$. Then for $Ysubseteq X$, clearly $sum_{xin Y}f(x)le sum _{xin X}f(x)$.
For $nin Bbb N$, let $X_n={,xin Xmid f(x)>frac 1n,}$. Then $$Mge sum_{xin X_n}f(x)ge sum_{xin X_n}frac 1n=frac 1n|X_n|$$
and in particular, $X_n$ must be finite. As $X=bigcup_{ninBbb N}X_n$, we conclude that $X$ is countable.
answered Nov 21 '18 at 18:05
Hagen von EitzenHagen von Eitzen
276k21269496
I think that the OP was talking about $I$ finite, in this case this is countable. Otherwise $S(I)$ doesn’t make sens.
– Thinking
Nov 21 '18 at 18:20
add a comment |
That sum, or just any sum with more than countably many positive summands cannot converge:
Let $X$ be a set and $fcolon XtoBbb R$ a map such that $f(x)> 0$ fore all $xin X$. Assume that $sum_{xin X}f(x)$ converges, say $sum_{xin X}f(x)=Sin Bbb R$. Then for $Ysubseteq X$, clearly $sum_{xin Y}f(x)le sum _{xin X}f(x)$.
For $nin Bbb N$, let $X_n={,xin Xmid f(x)>frac 1n,}$. Then $$Mge sum_{xin X_n}f(x)ge sum_{xin X_n}frac 1n=frac 1n|X_n|$$
and in particular, $X_n$ must be finite. As $X=bigcup_{ninBbb N}X_n$, we conclude that $X$ is countable.
answered Nov 21 '18 at 18:05
Hagen von EitzenHagen von Eitzen
276k21269496
I think that the OP was talking about $I$ finite, in this case this is countable. Otherwise $S(I)$ doesn’t make sens.
– Thinking
Nov 21 '18 at 18:20
add a comment |
That sum, or just any sum with more than countably many positive summands cannot converge:
Let $X$ be a set and $fcolon XtoBbb R$ a map such that $f(x)> 0$ fore all $xin X$. Assume that $sum_{xin X}f(x)$ converges, say $sum_{xin X}f(x)=Sin Bbb R$. Then for $Ysubseteq X$, clearly $sum_{xin Y}f(x)le sum _{xin X}f(x)$.
For $nin Bbb N$, let $X_n={,xin Xmid f(x)>frac 1n,}$. Then $$Mge sum_{xin X_n}f(x)ge sum_{xin X_n}frac 1n=frac 1n|X_n|$$
and in particular, $X_n$ must be finite. As $X=bigcup_{ninBbb N}X_n$, we conclude that $X$ is countable.
answered Nov 21 '18 at 18:05
Hagen von EitzenHagen von Eitzen
276k21269496
That sum, or just any sum with more than countably many positive summands cannot converge:
Let $X$ be a set and $fcolon XtoBbb R$ a map such that $f(x)> 0$ fore all $xin X$. Assume that $sum_{xin X}f(x)$ converges, say $sum_{xin X}f(x)=Sin Bbb R$. Then for $Ysubseteq X$, clearly $sum_{xin Y}f(x)le sum _{xin X}f(x)$.
For $nin Bbb N$, let $X_n={,xin Xmid f(x)>frac 1n,}$. Then $$Mge sum_{xin X_n}f(x)ge sum_{xin X_n}frac 1n=frac 1n|X_n|$$
and in particular, $X_n$ must be finite. As $X=bigcup_{ninBbb N}X_n$, we conclude that $X$ is countable.
answered Nov 21 '18 at 18:05
Hagen von EitzenHagen von Eitzen
276k21269496
answered Nov 21 '18 at 18:05
Hagen von EitzenHagen von Eitzen
276k21269496
answered Nov 21 '18 at 18:05
Hagen von EitzenHagen von Eitzen
276k21269496
answered Nov 21 '18 at 18:05
Hagen von EitzenHagen von Eitzen
276k21269496
276k21269496
I think that the OP was talking about $I$ finite, in this case this is countable. Otherwise $S(I)$ doesn’t make sens.
– Thinking
Nov 21 '18 at 18:20
add a comment |
I think that the OP was talking about $I$ finite, in this case this is countable. Otherwise $S(I)$ doesn’t make sens.
– Thinking
Nov 21 '18 at 18:20
I think that the OP was talking about $I$ finite, in this case this is countable. Otherwise $S(I)$ doesn’t make sens.
– Thinking
Nov 21 '18 at 18:20
I think that the OP was talking about $I$ finite, in this case this is countable. Otherwise $S(I)$ doesn’t make sens.
– Thinking
Nov 21 '18 at 18:20
add a comment |
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Can you confirm that $mid I mid$ is in $mathbb{R}$, I think it’s since otherwise $S(I)$ doesn’t make sens, but it’s just to be sure.
– Thinking
Nov 21 '18 at 18:22
Yes, $I < infty$.
– Interesting problems
Nov 21 '18 at 18:49