Mixing
Splitting infinite products up in a even and a odd part
Clash Royale CLAN TAG#URR8PPP
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While working on infinite products I was faced with the task to proof
$$prod_k=2^inftyleft(1+frac(-1)^kkright)~=~1$$
which is quite easy by using the fact that a even and the following odd factor are cancelling each other out
$$left(1+frac(-1)^2k2kright)left(1+frac(-1)^2k+12k+1right)~=~left(frac2k+(1)^k2kright)left(frac(2k+1)-(1)^k2k+1right)~=~frac2k+12kfrac2k2k+1$$
where the last term always equals $1$.
So I asked myself whether it is allowed to rewrite the infinite product in terms of even and odd factors or not. To point it out are these two definitions
$$prod_k=2^inftyleft(frack+(-1)^kkright)~=~prod_k=1^inftyfrac2k+12kfrac2k2k+1$$
equal? My intuition says yes hence both products are going from $k=0$ till infinity and so it should be irrelevant but I am not sure how to justify this intuition by myself.
infinite-product
asked Aug 1 at 14:57
mrtaurho
650117
add a comment |Â
up vote
1
down vote
favorite
While working on infinite products I was faced with the task to proof
$$prod_k=2^inftyleft(1+frac(-1)^kkright)~=~1$$
which is quite easy by using the fact that a even and the following odd factor are cancelling each other out
$$left(1+frac(-1)^2k2kright)left(1+frac(-1)^2k+12k+1right)~=~left(frac2k+(1)^k2kright)left(frac(2k+1)-(1)^k2k+1right)~=~frac2k+12kfrac2k2k+1$$
where the last term always equals $1$.
So I asked myself whether it is allowed to rewrite the infinite product in terms of even and odd factors or not. To point it out are these two definitions
$$prod_k=2^inftyleft(frack+(-1)^kkright)~=~prod_k=1^inftyfrac2k+12kfrac2k2k+1$$
equal? My intuition says yes hence both products are going from $k=0$ till infinity and so it should be irrelevant but I am not sure how to justify this intuition by myself.
infinite-product
asked Aug 1 at 14:57
mrtaurho
650117
Do you know the rigorous definition of an infinite product?
– John Gowers
Aug 1 at 14:58
To be honest: I am not exactly sure if I understand it well enough to draw my wanted conclusion from it.
– mrtaurho
Aug 1 at 15:00
This is analogous to inserting parentheses in an infinite sum.
– saulspatz
Aug 1 at 15:17
1
I read the title as "splitting infinitives" til I realised this wasn't English.SE
– IanF1
Aug 1 at 20:36
$prod_n=1^infty (frac5 + 3(-1)^n4) = 2 cdot frac12 cdot 2 cdot frac12 cdots$ does not converge even though grouping each pair of factors would give $1 cdot 1 cdot 1 cdots$ which does converge.
– Daniel Schepler
Aug 1 at 22:52
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
While working on infinite products I was faced with the task to proof
$$prod_k=2^inftyleft(1+frac(-1)^kkright)~=~1$$
which is quite easy by using the fact that a even and the following odd factor are cancelling each other out
$$left(1+frac(-1)^2k2kright)left(1+frac(-1)^2k+12k+1right)~=~left(frac2k+(1)^k2kright)left(frac(2k+1)-(1)^k2k+1right)~=~frac2k+12kfrac2k2k+1$$
where the last term always equals $1$.
So I asked myself whether it is allowed to rewrite the infinite product in terms of even and odd factors or not. To point it out are these two definitions
$$prod_k=2^inftyleft(frack+(-1)^kkright)~=~prod_k=1^inftyfrac2k+12kfrac2k2k+1$$
equal? My intuition says yes hence both products are going from $k=0$ till infinity and so it should be irrelevant but I am not sure how to justify this intuition by myself.
infinite-product
asked Aug 1 at 14:57
mrtaurho
650117
While working on infinite products I was faced with the task to proof
$$prod_k=2^inftyleft(1+frac(-1)^kkright)~=~1$$
which is quite easy by using the fact that a even and the following odd factor are cancelling each other out
$$left(1+frac(-1)^2k2kright)left(1+frac(-1)^2k+12k+1right)~=~left(frac2k+(1)^k2kright)left(frac(2k+1)-(1)^k2k+1right)~=~frac2k+12kfrac2k2k+1$$
where the last term always equals $1$.
So I asked myself whether it is allowed to rewrite the infinite product in terms of even and odd factors or not. To point it out are these two definitions
$$prod_k=2^inftyleft(frack+(-1)^kkright)~=~prod_k=1^inftyfrac2k+12kfrac2k2k+1$$
equal? My intuition says yes hence both products are going from $k=0$ till infinity and so it should be irrelevant but I am not sure how to justify this intuition by myself.
infinite-product
asked Aug 1 at 14:57
mrtaurho
650117
edited Aug 1 at 15:04
asked Aug 1 at 14:57
mrtaurho
650117
asked Aug 1 at 14:57
mrtaurho
650117
asked Aug 1 at 14:57
mrtaurho
650117
650117
Do you know the rigorous definition of an infinite product?
– John Gowers
Aug 1 at 14:58
To be honest: I am not exactly sure if I understand it well enough to draw my wanted conclusion from it.
– mrtaurho
Aug 1 at 15:00
This is analogous to inserting parentheses in an infinite sum.
– saulspatz
Aug 1 at 15:17
1
I read the title as "splitting infinitives" til I realised this wasn't English.SE
– IanF1
Aug 1 at 20:36
$prod_n=1^infty (frac5 + 3(-1)^n4) = 2 cdot frac12 cdot 2 cdot frac12 cdots$ does not converge even though grouping each pair of factors would give $1 cdot 1 cdot 1 cdots$ which does converge.
– Daniel Schepler
Aug 1 at 22:52
add a comment |Â
Do you know the rigorous definition of an infinite product?
– John Gowers
Aug 1 at 14:58
To be honest: I am not exactly sure if I understand it well enough to draw my wanted conclusion from it.
– mrtaurho
Aug 1 at 15:00
This is analogous to inserting parentheses in an infinite sum.
– saulspatz
Aug 1 at 15:17
1
I read the title as "splitting infinitives" til I realised this wasn't English.SE
– IanF1
Aug 1 at 20:36
$prod_n=1^infty (frac5 + 3(-1)^n4) = 2 cdot frac12 cdot 2 cdot frac12 cdots$ does not converge even though grouping each pair of factors would give $1 cdot 1 cdot 1 cdots$ which does converge.
– Daniel Schepler
Aug 1 at 22:52
Do you know the rigorous definition of an infinite product?
– John Gowers
Aug 1 at 14:58
Do you know the rigorous definition of an infinite product?
– John Gowers
Aug 1 at 14:58
To be honest: I am not exactly sure if I understand it well enough to draw my wanted conclusion from it.
– mrtaurho
Aug 1 at 15:00
To be honest: I am not exactly sure if I understand it well enough to draw my wanted conclusion from it.
– mrtaurho
Aug 1 at 15:00
This is analogous to inserting parentheses in an infinite sum.
– saulspatz
Aug 1 at 15:17
This is analogous to inserting parentheses in an infinite sum.
– saulspatz
Aug 1 at 15:17
1
1
I read the title as "splitting infinitives" til I realised this wasn't English.SE
– IanF1
Aug 1 at 20:36
I read the title as "splitting infinitives" til I realised this wasn't English.SE
– IanF1
Aug 1 at 20:36
$prod_n=1^infty (frac5 + 3(-1)^n4) = 2 cdot frac12 cdot 2 cdot frac12 cdots$ does not converge even though grouping each pair of factors would give $1 cdot 1 cdot 1 cdots$ which does converge.
– Daniel Schepler
Aug 1 at 22:52
$prod_n=1^infty (frac5 + 3(-1)^n4) = 2 cdot frac12 cdot 2 cdot frac12 cdots$ does not converge even though grouping each pair of factors would give $1 cdot 1 cdot 1 cdots$ which does converge.
– Daniel Schepler
Aug 1 at 22:52
add a comment |Â
3 Answers
3
active
oldest
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up vote
9
down vote
accepted
Note that the product $prod_k=2^2Kleft(1+frac(-1)^kkright)$ can be written in terms of even and odd terms as
$$beginalign
prod_k=2^2Kleft(1+frac(-1)^kkright)&=prod_k=1^K left(1+frac(-1)^2k2kright)prod_k=1^K-1 left(1+frac(-1)^2k+12k+1right)\\
&=prod_k=1^K left(frac2k+12kright)prod_k=1^K-1 left(frac2k2k+1right)\\
&=left(1+frac12Kright)prod_k=1^K-1 left(frac2k+12kfrac2k2k+1right)\\
&=left(1+frac12Kright)tag1
endalign$$
while the product $prod_k=2^2K+1left(1+frac(-1)^kkright)$ can be written
$$beginalign
prod_k=2^2K+1left(1+frac(-1)^kkright)&=prod_k=1^K left(1+frac(-1)^2k2kright)prod_k=1^K left(1+frac(-1)^2k+12k+1right)\\
&=prod_k=1^K left(frac2k+12kright)prod_k=1^K left(frac2k2k+1right)\\
&=prod_k=1^K left(frac2k+12kfrac2k2k+1right)\\
&=1tag2
endalign$$
Inasmuch as the right-hand sides of both $(1)$ and $(2)$ approach $1$ as $Ktoinfty$, we conclude
$$prod_k=2^inftyleft(1+frac(-1)^kkright)=1$$
as was to be shown!
answered Aug 1 at 15:21
Mark Viola
126k1171166
While this is correct, I do believe OP was looking for a pointer, rather than a whole explanation
– Rushabh Mehta
Aug 1 at 15:37
If you take for granted that the infinite product converges, you do not need to deal with the two cases separately.
– tomasz
Aug 1 at 18:37
1
@tomasz Indeed. For pedagogical purpose, I thought this presentation was appropriate.
– Mark Viola
Aug 1 at 20:49
Yeah. I was just remarking that, not saying you can take that for granted.
– tomasz
Aug 2 at 0:24
add a comment |Â
up vote
7
down vote
This is not true for general infinite products. If we define $a_2n=10^n$ and $a_2n+1= 10^-n$, then the infinite product
$$
prod_n=0^infty a_n
$$
is not defined, since the sequence of partial products is not convergent.
In your special case, however, you can get the result you want in this way. You'll need to be able to evaluate the finite product
$$
prod_k=2^N1+frac(-1)^kk,,
$$
(hint: you will likely need a different expression for odd and even $k$ - in fact, for even $k$, this product should come out to be $1$), and then show that that sequence converges to $1$ as $Ntoinfty$.
answered Aug 1 at 15:08
John Gowers
17.6k34168
add a comment |Â
up vote
1
down vote
Given
(1)$$f(2)f(3)f(4)f(5)...$$we can arrange the terms as
(2)$$[f(2)f(3)][f(4)f(5)]...$$ If we define $g(2k)=f(k)f(k+1)$, then (2) becomes
(3)$$g(2)g(4)...$$ If we define $j=2k$, then (3) becomes
(4)$$prod_j=1^inftyg(j)$$Since $k$ and $j$ are dummy variables, it doesn't matters what we use; $prod_j=1^inftyg(j)$ is the same as $prod_k=1^inftyg(k)$. In your case, $g(k) = 1$, so you conclude that the limit is 1.
There is, however, a complication. Going from (1) to (2) used the associative property of multiplication, which hold for finite products. One does, however, have to be careful about using it in infinite products. An example of whether the associative property doesn't hold is $(1-1)+(1-1)+(1-1)... neq 1+(-1+1)+(-1+1)...$. That's with sums, but a similar problem occurs with products. Basically, the problem is that you're taking partial products after an even number of terms, which means that you are getting a subsequence of partial products. If the subsequence of partial products, and the entire series, both have limits, then those limits are equal. But the subsequence having a limit does not mean that the entire sequence has a limit. So to be rigorous, you would first have to show that your infinite product has a limit, then use your argument to say that the limit is 1.
answered Aug 1 at 22:27
Acccumulation
4,0851314
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
Note that the product $prod_k=2^2Kleft(1+frac(-1)^kkright)$ can be written in terms of even and odd terms as
$$beginalign
prod_k=2^2Kleft(1+frac(-1)^kkright)&=prod_k=1^K left(1+frac(-1)^2k2kright)prod_k=1^K-1 left(1+frac(-1)^2k+12k+1right)\\
&=prod_k=1^K left(frac2k+12kright)prod_k=1^K-1 left(frac2k2k+1right)\\
&=left(1+frac12Kright)prod_k=1^K-1 left(frac2k+12kfrac2k2k+1right)\\
&=left(1+frac12Kright)tag1
endalign$$
while the product $prod_k=2^2K+1left(1+frac(-1)^kkright)$ can be written
$$beginalign
prod_k=2^2K+1left(1+frac(-1)^kkright)&=prod_k=1^K left(1+frac(-1)^2k2kright)prod_k=1^K left(1+frac(-1)^2k+12k+1right)\\
&=prod_k=1^K left(frac2k+12kright)prod_k=1^K left(frac2k2k+1right)\\
&=prod_k=1^K left(frac2k+12kfrac2k2k+1right)\\
&=1tag2
endalign$$
Inasmuch as the right-hand sides of both $(1)$ and $(2)$ approach $1$ as $Ktoinfty$, we conclude
$$prod_k=2^inftyleft(1+frac(-1)^kkright)=1$$
as was to be shown!
answered Aug 1 at 15:21
Mark Viola
126k1171166
While this is correct, I do believe OP was looking for a pointer, rather than a whole explanation
– Rushabh Mehta
Aug 1 at 15:37
If you take for granted that the infinite product converges, you do not need to deal with the two cases separately.
– tomasz
Aug 1 at 18:37
1
@tomasz Indeed. For pedagogical purpose, I thought this presentation was appropriate.
– Mark Viola
Aug 1 at 20:49
Yeah. I was just remarking that, not saying you can take that for granted.
– tomasz
Aug 2 at 0:24
add a comment |Â
up vote
9
down vote
accepted
Note that the product $prod_k=2^2Kleft(1+frac(-1)^kkright)$ can be written in terms of even and odd terms as
$$beginalign
prod_k=2^2Kleft(1+frac(-1)^kkright)&=prod_k=1^K left(1+frac(-1)^2k2kright)prod_k=1^K-1 left(1+frac(-1)^2k+12k+1right)\\
&=prod_k=1^K left(frac2k+12kright)prod_k=1^K-1 left(frac2k2k+1right)\\
&=left(1+frac12Kright)prod_k=1^K-1 left(frac2k+12kfrac2k2k+1right)\\
&=left(1+frac12Kright)tag1
endalign$$
while the product $prod_k=2^2K+1left(1+frac(-1)^kkright)$ can be written
$$beginalign
prod_k=2^2K+1left(1+frac(-1)^kkright)&=prod_k=1^K left(1+frac(-1)^2k2kright)prod_k=1^K left(1+frac(-1)^2k+12k+1right)\\
&=prod_k=1^K left(frac2k+12kright)prod_k=1^K left(frac2k2k+1right)\\
&=prod_k=1^K left(frac2k+12kfrac2k2k+1right)\\
&=1tag2
endalign$$
Inasmuch as the right-hand sides of both $(1)$ and $(2)$ approach $1$ as $Ktoinfty$, we conclude
$$prod_k=2^inftyleft(1+frac(-1)^kkright)=1$$
as was to be shown!
answered Aug 1 at 15:21
Mark Viola
126k1171166
While this is correct, I do believe OP was looking for a pointer, rather than a whole explanation
– Rushabh Mehta
Aug 1 at 15:37
If you take for granted that the infinite product converges, you do not need to deal with the two cases separately.
– tomasz
Aug 1 at 18:37
1
@tomasz Indeed. For pedagogical purpose, I thought this presentation was appropriate.
– Mark Viola
Aug 1 at 20:49
Yeah. I was just remarking that, not saying you can take that for granted.
– tomasz
Aug 2 at 0:24
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
Note that the product $prod_k=2^2Kleft(1+frac(-1)^kkright)$ can be written in terms of even and odd terms as
$$beginalign
prod_k=2^2Kleft(1+frac(-1)^kkright)&=prod_k=1^K left(1+frac(-1)^2k2kright)prod_k=1^K-1 left(1+frac(-1)^2k+12k+1right)\\
&=prod_k=1^K left(frac2k+12kright)prod_k=1^K-1 left(frac2k2k+1right)\\
&=left(1+frac12Kright)prod_k=1^K-1 left(frac2k+12kfrac2k2k+1right)\\
&=left(1+frac12Kright)tag1
endalign$$
while the product $prod_k=2^2K+1left(1+frac(-1)^kkright)$ can be written
$$beginalign
prod_k=2^2K+1left(1+frac(-1)^kkright)&=prod_k=1^K left(1+frac(-1)^2k2kright)prod_k=1^K left(1+frac(-1)^2k+12k+1right)\\
&=prod_k=1^K left(frac2k+12kright)prod_k=1^K left(frac2k2k+1right)\\
&=prod_k=1^K left(frac2k+12kfrac2k2k+1right)\\
&=1tag2
endalign$$
Inasmuch as the right-hand sides of both $(1)$ and $(2)$ approach $1$ as $Ktoinfty$, we conclude
$$prod_k=2^inftyleft(1+frac(-1)^kkright)=1$$
as was to be shown!
answered Aug 1 at 15:21
Mark Viola
126k1171166
Note that the product $prod_k=2^2Kleft(1+frac(-1)^kkright)$ can be written in terms of even and odd terms as
$$beginalign
prod_k=2^2Kleft(1+frac(-1)^kkright)&=prod_k=1^K left(1+frac(-1)^2k2kright)prod_k=1^K-1 left(1+frac(-1)^2k+12k+1right)\\
&=prod_k=1^K left(frac2k+12kright)prod_k=1^K-1 left(frac2k2k+1right)\\
&=left(1+frac12Kright)prod_k=1^K-1 left(frac2k+12kfrac2k2k+1right)\\
&=left(1+frac12Kright)tag1
endalign$$
while the product $prod_k=2^2K+1left(1+frac(-1)^kkright)$ can be written
$$beginalign
prod_k=2^2K+1left(1+frac(-1)^kkright)&=prod_k=1^K left(1+frac(-1)^2k2kright)prod_k=1^K left(1+frac(-1)^2k+12k+1right)\\
&=prod_k=1^K left(frac2k+12kright)prod_k=1^K left(frac2k2k+1right)\\
&=prod_k=1^K left(frac2k+12kfrac2k2k+1right)\\
&=1tag2
endalign$$
Inasmuch as the right-hand sides of both $(1)$ and $(2)$ approach $1$ as $Ktoinfty$, we conclude
$$prod_k=2^inftyleft(1+frac(-1)^kkright)=1$$
as was to be shown!
answered Aug 1 at 15:21
Mark Viola
126k1171166
answered Aug 1 at 15:21
Mark Viola
126k1171166
answered Aug 1 at 15:21
Mark Viola
126k1171166
answered Aug 1 at 15:21
Mark Viola
126k1171166
126k1171166
While this is correct, I do believe OP was looking for a pointer, rather than a whole explanation
– Rushabh Mehta
Aug 1 at 15:37
If you take for granted that the infinite product converges, you do not need to deal with the two cases separately.
– tomasz
Aug 1 at 18:37
1
@tomasz Indeed. For pedagogical purpose, I thought this presentation was appropriate.
– Mark Viola
Aug 1 at 20:49
Yeah. I was just remarking that, not saying you can take that for granted.
– tomasz
Aug 2 at 0:24
add a comment |Â
While this is correct, I do believe OP was looking for a pointer, rather than a whole explanation
– Rushabh Mehta
Aug 1 at 15:37
If you take for granted that the infinite product converges, you do not need to deal with the two cases separately.
– tomasz
Aug 1 at 18:37
1
@tomasz Indeed. For pedagogical purpose, I thought this presentation was appropriate.
– Mark Viola
Aug 1 at 20:49
Yeah. I was just remarking that, not saying you can take that for granted.
– tomasz
Aug 2 at 0:24
While this is correct, I do believe OP was looking for a pointer, rather than a whole explanation
– Rushabh Mehta
Aug 1 at 15:37
While this is correct, I do believe OP was looking for a pointer, rather than a whole explanation
– Rushabh Mehta
Aug 1 at 15:37
If you take for granted that the infinite product converges, you do not need to deal with the two cases separately.
– tomasz
Aug 1 at 18:37
If you take for granted that the infinite product converges, you do not need to deal with the two cases separately.
– tomasz
Aug 1 at 18:37
1
1
@tomasz Indeed. For pedagogical purpose, I thought this presentation was appropriate.
– Mark Viola
Aug 1 at 20:49
@tomasz Indeed. For pedagogical purpose, I thought this presentation was appropriate.
– Mark Viola
Aug 1 at 20:49
Yeah. I was just remarking that, not saying you can take that for granted.
– tomasz
Aug 2 at 0:24
Yeah. I was just remarking that, not saying you can take that for granted.
– tomasz
Aug 2 at 0:24
add a comment |Â
up vote
7
down vote
This is not true for general infinite products. If we define $a_2n=10^n$ and $a_2n+1= 10^-n$, then the infinite product
$$
prod_n=0^infty a_n
$$
is not defined, since the sequence of partial products is not convergent.
In your special case, however, you can get the result you want in this way. You'll need to be able to evaluate the finite product
$$
prod_k=2^N1+frac(-1)^kk,,
$$
(hint: you will likely need a different expression for odd and even $k$ - in fact, for even $k$, this product should come out to be $1$), and then show that that sequence converges to $1$ as $Ntoinfty$.
answered Aug 1 at 15:08
John Gowers
17.6k34168
add a comment |Â
up vote
7
down vote
This is not true for general infinite products. If we define $a_2n=10^n$ and $a_2n+1= 10^-n$, then the infinite product
$$
prod_n=0^infty a_n
$$
is not defined, since the sequence of partial products is not convergent.
In your special case, however, you can get the result you want in this way. You'll need to be able to evaluate the finite product
$$
prod_k=2^N1+frac(-1)^kk,,
$$
(hint: you will likely need a different expression for odd and even $k$ - in fact, for even $k$, this product should come out to be $1$), and then show that that sequence converges to $1$ as $Ntoinfty$.
answered Aug 1 at 15:08
John Gowers
17.6k34168
add a comment |Â
up vote
7
down vote
up vote
7
down vote
This is not true for general infinite products. If we define $a_2n=10^n$ and $a_2n+1= 10^-n$, then the infinite product
$$
prod_n=0^infty a_n
$$
is not defined, since the sequence of partial products is not convergent.
In your special case, however, you can get the result you want in this way. You'll need to be able to evaluate the finite product
$$
prod_k=2^N1+frac(-1)^kk,,
$$
(hint: you will likely need a different expression for odd and even $k$ - in fact, for even $k$, this product should come out to be $1$), and then show that that sequence converges to $1$ as $Ntoinfty$.
answered Aug 1 at 15:08
John Gowers
17.6k34168
This is not true for general infinite products. If we define $a_2n=10^n$ and $a_2n+1= 10^-n$, then the infinite product
$$
prod_n=0^infty a_n
$$
is not defined, since the sequence of partial products is not convergent.
In your special case, however, you can get the result you want in this way. You'll need to be able to evaluate the finite product
$$
prod_k=2^N1+frac(-1)^kk,,
$$
(hint: you will likely need a different expression for odd and even $k$ - in fact, for even $k$, this product should come out to be $1$), and then show that that sequence converges to $1$ as $Ntoinfty$.
answered Aug 1 at 15:08
John Gowers
17.6k34168
answered Aug 1 at 15:08
John Gowers
17.6k34168
answered Aug 1 at 15:08
John Gowers
17.6k34168
answered Aug 1 at 15:08
John Gowers
17.6k34168
17.6k34168
add a comment |Â
add a comment |Â
up vote
1
down vote
Given
(1)$$f(2)f(3)f(4)f(5)...$$we can arrange the terms as
(2)$$[f(2)f(3)][f(4)f(5)]...$$ If we define $g(2k)=f(k)f(k+1)$, then (2) becomes
(3)$$g(2)g(4)...$$ If we define $j=2k$, then (3) becomes
(4)$$prod_j=1^inftyg(j)$$Since $k$ and $j$ are dummy variables, it doesn't matters what we use; $prod_j=1^inftyg(j)$ is the same as $prod_k=1^inftyg(k)$. In your case, $g(k) = 1$, so you conclude that the limit is 1.
There is, however, a complication. Going from (1) to (2) used the associative property of multiplication, which hold for finite products. One does, however, have to be careful about using it in infinite products. An example of whether the associative property doesn't hold is $(1-1)+(1-1)+(1-1)... neq 1+(-1+1)+(-1+1)...$. That's with sums, but a similar problem occurs with products. Basically, the problem is that you're taking partial products after an even number of terms, which means that you are getting a subsequence of partial products. If the subsequence of partial products, and the entire series, both have limits, then those limits are equal. But the subsequence having a limit does not mean that the entire sequence has a limit. So to be rigorous, you would first have to show that your infinite product has a limit, then use your argument to say that the limit is 1.
answered Aug 1 at 22:27
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Given
(1)$$f(2)f(3)f(4)f(5)...$$we can arrange the terms as
(2)$$[f(2)f(3)][f(4)f(5)]...$$ If we define $g(2k)=f(k)f(k+1)$, then (2) becomes
(3)$$g(2)g(4)...$$ If we define $j=2k$, then (3) becomes
(4)$$prod_j=1^inftyg(j)$$Since $k$ and $j$ are dummy variables, it doesn't matters what we use; $prod_j=1^inftyg(j)$ is the same as $prod_k=1^inftyg(k)$. In your case, $g(k) = 1$, so you conclude that the limit is 1.
There is, however, a complication. Going from (1) to (2) used the associative property of multiplication, which hold for finite products. One does, however, have to be careful about using it in infinite products. An example of whether the associative property doesn't hold is $(1-1)+(1-1)+(1-1)... neq 1+(-1+1)+(-1+1)...$. That's with sums, but a similar problem occurs with products. Basically, the problem is that you're taking partial products after an even number of terms, which means that you are getting a subsequence of partial products. If the subsequence of partial products, and the entire series, both have limits, then those limits are equal. But the subsequence having a limit does not mean that the entire sequence has a limit. So to be rigorous, you would first have to show that your infinite product has a limit, then use your argument to say that the limit is 1.
answered Aug 1 at 22:27
Acccumulation
4,0851314
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Given
(1)$$f(2)f(3)f(4)f(5)...$$we can arrange the terms as
(2)$$[f(2)f(3)][f(4)f(5)]...$$ If we define $g(2k)=f(k)f(k+1)$, then (2) becomes
(3)$$g(2)g(4)...$$ If we define $j=2k$, then (3) becomes
(4)$$prod_j=1^inftyg(j)$$Since $k$ and $j$ are dummy variables, it doesn't matters what we use; $prod_j=1^inftyg(j)$ is the same as $prod_k=1^inftyg(k)$. In your case, $g(k) = 1$, so you conclude that the limit is 1.
There is, however, a complication. Going from (1) to (2) used the associative property of multiplication, which hold for finite products. One does, however, have to be careful about using it in infinite products. An example of whether the associative property doesn't hold is $(1-1)+(1-1)+(1-1)... neq 1+(-1+1)+(-1+1)...$. That's with sums, but a similar problem occurs with products. Basically, the problem is that you're taking partial products after an even number of terms, which means that you are getting a subsequence of partial products. If the subsequence of partial products, and the entire series, both have limits, then those limits are equal. But the subsequence having a limit does not mean that the entire sequence has a limit. So to be rigorous, you would first have to show that your infinite product has a limit, then use your argument to say that the limit is 1.
answered Aug 1 at 22:27
Acccumulation
4,0851314
Given
(1)$$f(2)f(3)f(4)f(5)...$$we can arrange the terms as
(2)$$[f(2)f(3)][f(4)f(5)]...$$ If we define $g(2k)=f(k)f(k+1)$, then (2) becomes
(3)$$g(2)g(4)...$$ If we define $j=2k$, then (3) becomes
(4)$$prod_j=1^inftyg(j)$$Since $k$ and $j$ are dummy variables, it doesn't matters what we use; $prod_j=1^inftyg(j)$ is the same as $prod_k=1^inftyg(k)$. In your case, $g(k) = 1$, so you conclude that the limit is 1.
There is, however, a complication. Going from (1) to (2) used the associative property of multiplication, which hold for finite products. One does, however, have to be careful about using it in infinite products. An example of whether the associative property doesn't hold is $(1-1)+(1-1)+(1-1)... neq 1+(-1+1)+(-1+1)...$. That's with sums, but a similar problem occurs with products. Basically, the problem is that you're taking partial products after an even number of terms, which means that you are getting a subsequence of partial products. If the subsequence of partial products, and the entire series, both have limits, then those limits are equal. But the subsequence having a limit does not mean that the entire sequence has a limit. So to be rigorous, you would first have to show that your infinite product has a limit, then use your argument to say that the limit is 1.
answered Aug 1 at 22:27
Acccumulation
4,0851314
answered Aug 1 at 22:27
Acccumulation
4,0851314
answered Aug 1 at 22:27
Acccumulation
4,0851314
answered Aug 1 at 22:27
Acccumulation
4,0851314
4,0851314
add a comment |Â
add a comment |Â
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Do you know the rigorous definition of an infinite product?
– John Gowers
Aug 1 at 14:58
To be honest: I am not exactly sure if I understand it well enough to draw my wanted conclusion from it.
– mrtaurho
Aug 1 at 15:00
This is analogous to inserting parentheses in an infinite sum.
– saulspatz
Aug 1 at 15:17
1
I read the title as "splitting infinitives" til I realised this wasn't English.SE
– IanF1
Aug 1 at 20:36
$prod_n=1^infty (frac5 + 3(-1)^n4) = 2 cdot frac12 cdot 2 cdot frac12 cdots$ does not converge even though grouping each pair of factors would give $1 cdot 1 cdot 1 cdots$ which does converge.
– Daniel Schepler
Aug 1 at 22:52