Mixing
Interchanging the order of infinite and finite sum
Clash Royale CLAN TAG#URR8PPP
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I have the following series
$$fleft( k,t right) = sumlimits_n = 0^infty a_n(k)t^n.$$
It is obvious that
$$sumlimits_k = 0^m fleft( k,t right) = sumlimits_k = 0^m sumlimits_n = 0^infty a_n(k)t^n .$$
But I don't know if the following is true or not
$$sumlimits_k = 0^m sumlimits_n = 0^infty a_n(k)t^n = sumlimits_n = 0^infty sumlimits_k = 0^m a_n(k)t^n $$
I will appreciate if someone tell me under what conditions I can change the order of this double series.
sequences-and-series
asked Jul 25 at 6:51
p.kn
1087
add a comment |Â
up vote
2
down vote
favorite
I have the following series
$$fleft( k,t right) = sumlimits_n = 0^infty a_n(k)t^n.$$
It is obvious that
$$sumlimits_k = 0^m fleft( k,t right) = sumlimits_k = 0^m sumlimits_n = 0^infty a_n(k)t^n .$$
But I don't know if the following is true or not
$$sumlimits_k = 0^m sumlimits_n = 0^infty a_n(k)t^n = sumlimits_n = 0^infty sumlimits_k = 0^m a_n(k)t^n $$
I will appreciate if someone tell me under what conditions I can change the order of this double series.
sequences-and-series
asked Jul 25 at 6:51
p.kn
1087
1
I think absolute convergence is sufficient, but maybe this condition can be relaxed.
– Szeto
Jul 25 at 7:00
1
You can use Fubini's theorem with counting measures
– Cuoredicervo
Jul 25 at 7:10
Well it is not in general true, but if you know the double sum is finite, you can use Fubini's theorem with the counting measure to obtain your claim (well it works almost everywhere, but I guess that's fine). If you want it to work everywhere, you probably need absolute convergence.
– Raymond Chu
Jul 25 at 7:33
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have the following series
$$fleft( k,t right) = sumlimits_n = 0^infty a_n(k)t^n.$$
It is obvious that
$$sumlimits_k = 0^m fleft( k,t right) = sumlimits_k = 0^m sumlimits_n = 0^infty a_n(k)t^n .$$
But I don't know if the following is true or not
$$sumlimits_k = 0^m sumlimits_n = 0^infty a_n(k)t^n = sumlimits_n = 0^infty sumlimits_k = 0^m a_n(k)t^n $$
I will appreciate if someone tell me under what conditions I can change the order of this double series.
sequences-and-series
asked Jul 25 at 6:51
p.kn
1087
I have the following series
$$fleft( k,t right) = sumlimits_n = 0^infty a_n(k)t^n.$$
It is obvious that
$$sumlimits_k = 0^m fleft( k,t right) = sumlimits_k = 0^m sumlimits_n = 0^infty a_n(k)t^n .$$
But I don't know if the following is true or not
$$sumlimits_k = 0^m sumlimits_n = 0^infty a_n(k)t^n = sumlimits_n = 0^infty sumlimits_k = 0^m a_n(k)t^n $$
I will appreciate if someone tell me under what conditions I can change the order of this double series.
sequences-and-series
asked Jul 25 at 6:51
p.kn
1087
asked Jul 25 at 6:51
p.kn
1087
asked Jul 25 at 6:51
p.kn
1087
asked Jul 25 at 6:51
p.kn
1087
1087
1
I think absolute convergence is sufficient, but maybe this condition can be relaxed.
– Szeto
Jul 25 at 7:00
1
You can use Fubini's theorem with counting measures
– Cuoredicervo
Jul 25 at 7:10
Well it is not in general true, but if you know the double sum is finite, you can use Fubini's theorem with the counting measure to obtain your claim (well it works almost everywhere, but I guess that's fine). If you want it to work everywhere, you probably need absolute convergence.
– Raymond Chu
Jul 25 at 7:33
add a comment |Â
1
I think absolute convergence is sufficient, but maybe this condition can be relaxed.
– Szeto
Jul 25 at 7:00
1
You can use Fubini's theorem with counting measures
– Cuoredicervo
Jul 25 at 7:10
Well it is not in general true, but if you know the double sum is finite, you can use Fubini's theorem with the counting measure to obtain your claim (well it works almost everywhere, but I guess that's fine). If you want it to work everywhere, you probably need absolute convergence.
– Raymond Chu
Jul 25 at 7:33
1
1
I think absolute convergence is sufficient, but maybe this condition can be relaxed.
– Szeto
Jul 25 at 7:00
I think absolute convergence is sufficient, but maybe this condition can be relaxed.
– Szeto
Jul 25 at 7:00
1
1
You can use Fubini's theorem with counting measures
– Cuoredicervo
Jul 25 at 7:10
You can use Fubini's theorem with counting measures
– Cuoredicervo
Jul 25 at 7:10
Well it is not in general true, but if you know the double sum is finite, you can use Fubini's theorem with the counting measure to obtain your claim (well it works almost everywhere, but I guess that's fine). If you want it to work everywhere, you probably need absolute convergence.
– Raymond Chu
Jul 25 at 7:33
Well it is not in general true, but if you know the double sum is finite, you can use Fubini's theorem with the counting measure to obtain your claim (well it works almost everywhere, but I guess that's fine). If you want it to work everywhere, you probably need absolute convergence.
– Raymond Chu
Jul 25 at 7:33
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
That is not always true.
Take $t=1$, $m=1$, $a_n(0)= (-1)^n$ and $a_n(1)=(-1)^n+1$. The equality
$$sumlimits_k = 0^m sumlimits_n = 0^infty a_n(k)t^n = sumlimits_n = 0^infty sumlimits_k = 0^m a_n(k)t^n $$ is not satisfied as $sum (-1)^n$ doesn't converge.
answered Jul 25 at 7:11
mathcounterexamples.net
23.8k21652
thank you. But in my case the infinite series equals a known two variable function, and I want to change the order of infinite series with finite. actually I want to know under what conditions I am allowed to do this.
– p.kn
Jul 25 at 8:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
That is not always true.
Take $t=1$, $m=1$, $a_n(0)= (-1)^n$ and $a_n(1)=(-1)^n+1$. The equality
$$sumlimits_k = 0^m sumlimits_n = 0^infty a_n(k)t^n = sumlimits_n = 0^infty sumlimits_k = 0^m a_n(k)t^n $$ is not satisfied as $sum (-1)^n$ doesn't converge.
answered Jul 25 at 7:11
mathcounterexamples.net
23.8k21652
thank you. But in my case the infinite series equals a known two variable function, and I want to change the order of infinite series with finite. actually I want to know under what conditions I am allowed to do this.
– p.kn
Jul 25 at 8:03
add a comment |Â
up vote
1
down vote
That is not always true.
Take $t=1$, $m=1$, $a_n(0)= (-1)^n$ and $a_n(1)=(-1)^n+1$. The equality
$$sumlimits_k = 0^m sumlimits_n = 0^infty a_n(k)t^n = sumlimits_n = 0^infty sumlimits_k = 0^m a_n(k)t^n $$ is not satisfied as $sum (-1)^n$ doesn't converge.
answered Jul 25 at 7:11
mathcounterexamples.net
23.8k21652
thank you. But in my case the infinite series equals a known two variable function, and I want to change the order of infinite series with finite. actually I want to know under what conditions I am allowed to do this.
– p.kn
Jul 25 at 8:03
add a comment |Â
up vote
1
down vote
up vote
1
down vote
That is not always true.
Take $t=1$, $m=1$, $a_n(0)= (-1)^n$ and $a_n(1)=(-1)^n+1$. The equality
$$sumlimits_k = 0^m sumlimits_n = 0^infty a_n(k)t^n = sumlimits_n = 0^infty sumlimits_k = 0^m a_n(k)t^n $$ is not satisfied as $sum (-1)^n$ doesn't converge.
answered Jul 25 at 7:11
mathcounterexamples.net
23.8k21652
That is not always true.
Take $t=1$, $m=1$, $a_n(0)= (-1)^n$ and $a_n(1)=(-1)^n+1$. The equality
$$sumlimits_k = 0^m sumlimits_n = 0^infty a_n(k)t^n = sumlimits_n = 0^infty sumlimits_k = 0^m a_n(k)t^n $$ is not satisfied as $sum (-1)^n$ doesn't converge.
answered Jul 25 at 7:11
mathcounterexamples.net
23.8k21652
answered Jul 25 at 7:11
mathcounterexamples.net
23.8k21652
answered Jul 25 at 7:11
mathcounterexamples.net
23.8k21652
answered Jul 25 at 7:11
mathcounterexamples.net
23.8k21652
23.8k21652
thank you. But in my case the infinite series equals a known two variable function, and I want to change the order of infinite series with finite. actually I want to know under what conditions I am allowed to do this.
– p.kn
Jul 25 at 8:03
add a comment |Â
thank you. But in my case the infinite series equals a known two variable function, and I want to change the order of infinite series with finite. actually I want to know under what conditions I am allowed to do this.
– p.kn
Jul 25 at 8:03
thank you. But in my case the infinite series equals a known two variable function, and I want to change the order of infinite series with finite. actually I want to know under what conditions I am allowed to do this.
– p.kn
Jul 25 at 8:03
thank you. But in my case the infinite series equals a known two variable function, and I want to change the order of infinite series with finite. actually I want to know under what conditions I am allowed to do this.
– p.kn
Jul 25 at 8:03
add a comment |Â
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1
I think absolute convergence is sufficient, but maybe this condition can be relaxed.
– Szeto
Jul 25 at 7:00
1
You can use Fubini's theorem with counting measures
– Cuoredicervo
Jul 25 at 7:10
Well it is not in general true, but if you know the double sum is finite, you can use Fubini's theorem with the counting measure to obtain your claim (well it works almost everywhere, but I guess that's fine). If you want it to work everywhere, you probably need absolute convergence.
– Raymond Chu
Jul 25 at 7:33