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lower bound of convergent infinite products
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If an infinite product
[
prod_n (1 + a_n(s))
]
converges, then can we say anything about its lower bound?
Or in general, can it take any value arbitrarily close to $0$?
What if $a_n$ is of the form $a_n(s) = b_nm_n^-s$, where $m_n $ is a subsequence of $1, 2, ldots $ and |b_n| = 1?
calculus infinite-product
asked Jul 19 at 3:42
Grown pains
9510
 |Â
show 4 more comments
up vote
0
down vote
favorite
If an infinite product
[
prod_n (1 + a_n(s))
]
converges, then can we say anything about its lower bound?
Or in general, can it take any value arbitrarily close to $0$?
What if $a_n$ is of the form $a_n(s) = b_nm_n^-s$, where $m_n $ is a subsequence of $1, 2, ldots $ and |b_n| = 1?
calculus infinite-product
asked Jul 19 at 3:42
Grown pains
9510
I get the impression that you probably have conditions on the $a_n(s)$ in mind that you didn't state. As written, the $1+$ is irrelevant; you could just as well shift the $a_n(s)$ and write $prod_na_n(s)$, and you could trivially make the product come out as any desired value simply by changing the first factor accordingly.
– joriki
Jul 19 at 4:21
Then only condition is $a_n(s) to 0$ as $n to infty$.
– Grown pains
Jul 19 at 4:27
That's not an additional condition; that follows from the convergence of the product, which is already a condition in the question.
– joriki
Jul 19 at 4:36
Okay. What if $a_n(s)$ is of the form $c_nm_n^-s$ where $|c_n| = 1$ and $m_n$ is a subsequence of all naturals?
– Grown pains
Jul 19 at 4:43
The naturals aren't a sequence, so they don't have subsequences.
– joriki
Jul 19 at 4:48
 |Â
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If an infinite product
[
prod_n (1 + a_n(s))
]
converges, then can we say anything about its lower bound?
Or in general, can it take any value arbitrarily close to $0$?
What if $a_n$ is of the form $a_n(s) = b_nm_n^-s$, where $m_n $ is a subsequence of $1, 2, ldots $ and |b_n| = 1?
calculus infinite-product
asked Jul 19 at 3:42
Grown pains
9510
If an infinite product
[
prod_n (1 + a_n(s))
]
converges, then can we say anything about its lower bound?
Or in general, can it take any value arbitrarily close to $0$?
What if $a_n$ is of the form $a_n(s) = b_nm_n^-s$, where $m_n $ is a subsequence of $1, 2, ldots $ and |b_n| = 1?
calculus infinite-product
asked Jul 19 at 3:42
Grown pains
9510
edited Jul 19 at 8:40
asked Jul 19 at 3:42
Grown pains
9510
asked Jul 19 at 3:42
Grown pains
9510
asked Jul 19 at 3:42
Grown pains
9510
9510
I get the impression that you probably have conditions on the $a_n(s)$ in mind that you didn't state. As written, the $1+$ is irrelevant; you could just as well shift the $a_n(s)$ and write $prod_na_n(s)$, and you could trivially make the product come out as any desired value simply by changing the first factor accordingly.
– joriki
Jul 19 at 4:21
Then only condition is $a_n(s) to 0$ as $n to infty$.
– Grown pains
Jul 19 at 4:27
That's not an additional condition; that follows from the convergence of the product, which is already a condition in the question.
– joriki
Jul 19 at 4:36
Okay. What if $a_n(s)$ is of the form $c_nm_n^-s$ where $|c_n| = 1$ and $m_n$ is a subsequence of all naturals?
– Grown pains
Jul 19 at 4:43
The naturals aren't a sequence, so they don't have subsequences.
– joriki
Jul 19 at 4:48
 |Â
show 4 more comments
I get the impression that you probably have conditions on the $a_n(s)$ in mind that you didn't state. As written, the $1+$ is irrelevant; you could just as well shift the $a_n(s)$ and write $prod_na_n(s)$, and you could trivially make the product come out as any desired value simply by changing the first factor accordingly.
– joriki
Jul 19 at 4:21
Then only condition is $a_n(s) to 0$ as $n to infty$.
– Grown pains
Jul 19 at 4:27
That's not an additional condition; that follows from the convergence of the product, which is already a condition in the question.
– joriki
Jul 19 at 4:36
Okay. What if $a_n(s)$ is of the form $c_nm_n^-s$ where $|c_n| = 1$ and $m_n$ is a subsequence of all naturals?
– Grown pains
Jul 19 at 4:43
The naturals aren't a sequence, so they don't have subsequences.
– joriki
Jul 19 at 4:48
I get the impression that you probably have conditions on the $a_n(s)$ in mind that you didn't state. As written, the $1+$ is irrelevant; you could just as well shift the $a_n(s)$ and write $prod_na_n(s)$, and you could trivially make the product come out as any desired value simply by changing the first factor accordingly.
– joriki
Jul 19 at 4:21
I get the impression that you probably have conditions on the $a_n(s)$ in mind that you didn't state. As written, the $1+$ is irrelevant; you could just as well shift the $a_n(s)$ and write $prod_na_n(s)$, and you could trivially make the product come out as any desired value simply by changing the first factor accordingly.
– joriki
Jul 19 at 4:21
Then only condition is $a_n(s) to 0$ as $n to infty$.
– Grown pains
Jul 19 at 4:27
Then only condition is $a_n(s) to 0$ as $n to infty$.
– Grown pains
Jul 19 at 4:27
That's not an additional condition; that follows from the convergence of the product, which is already a condition in the question.
– joriki
Jul 19 at 4:36
That's not an additional condition; that follows from the convergence of the product, which is already a condition in the question.
– joriki
Jul 19 at 4:36
Okay. What if $a_n(s)$ is of the form $c_nm_n^-s$ where $|c_n| = 1$ and $m_n$ is a subsequence of all naturals?
– Grown pains
Jul 19 at 4:43
Okay. What if $a_n(s)$ is of the form $c_nm_n^-s$ where $|c_n| = 1$ and $m_n$ is a subsequence of all naturals?
– Grown pains
Jul 19 at 4:43
The naturals aren't a sequence, so they don't have subsequences.
– joriki
Jul 19 at 4:48
The naturals aren't a sequence, so they don't have subsequences.
– joriki
Jul 19 at 4:48
 |Â
show 4 more comments
1 Answer
1
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oldest
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up vote
1
down vote
The product can take an arbitrary value, in particular any value arbitrarily close to zero. Take any infinite product that converges to some value $alpha$. To let it converge to arbitrary $betainmathbb C$ instead, multiply the first factor by $beta/alpha$. In terms of your factors $1+a_n(s)$, you need
$$
1+a_n'(s)=fracbetaalpha(1+a_n(s))
$$
and thus
$$
a_n'(s)=fracbetaalpha(1+a_n(s))-1;.
$$
Edit in response to the new question:
The product can still be zero if $b_1=-1$ and $m_1=1$. Excluding this trivial case, whether the product can converge to zero now depends on $s$.
Since the product converges, $a_n$ converges to $0$; it follows that $sgt0$. Then the product is minimal if $b_n=-1$ and $m_n=n$ for all $nge2$, yielding
$$
prod_nge2left(1-n^-sright);,
$$
with another factor of $2$ if we choose to include $m_1=1$ with $b_1=1$. We can obtain a lower bound as follows:
begineqnarray*
prod_nge2left(1-n^-sright)
&=&
expleft(sum_nge2logleft(1-n^-sright)right)\
<&
expleft(-sum_nge2n^-sright)\
&=&expleft(1-zeta(s)right);.
endeqnarray*
This holds for $sgt1$. The zeta function has a pole at $s=1$, and in this case the sum in the exponent diverges logarithmically as $ntoinfty$. Thus, for $sle1$ the product can converge to zero.
answered Jul 19 at 4:40
joriki
164k10180328
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
The product can take an arbitrary value, in particular any value arbitrarily close to zero. Take any infinite product that converges to some value $alpha$. To let it converge to arbitrary $betainmathbb C$ instead, multiply the first factor by $beta/alpha$. In terms of your factors $1+a_n(s)$, you need
$$
1+a_n'(s)=fracbetaalpha(1+a_n(s))
$$
and thus
$$
a_n'(s)=fracbetaalpha(1+a_n(s))-1;.
$$
Edit in response to the new question:
The product can still be zero if $b_1=-1$ and $m_1=1$. Excluding this trivial case, whether the product can converge to zero now depends on $s$.
Since the product converges, $a_n$ converges to $0$; it follows that $sgt0$. Then the product is minimal if $b_n=-1$ and $m_n=n$ for all $nge2$, yielding
$$
prod_nge2left(1-n^-sright);,
$$
with another factor of $2$ if we choose to include $m_1=1$ with $b_1=1$. We can obtain a lower bound as follows:
begineqnarray*
prod_nge2left(1-n^-sright)
&=&
expleft(sum_nge2logleft(1-n^-sright)right)\
<&
expleft(-sum_nge2n^-sright)\
&=&expleft(1-zeta(s)right);.
endeqnarray*
This holds for $sgt1$. The zeta function has a pole at $s=1$, and in this case the sum in the exponent diverges logarithmically as $ntoinfty$. Thus, for $sle1$ the product can converge to zero.
answered Jul 19 at 4:40
joriki
164k10180328
add a comment |Â
up vote
1
down vote
The product can take an arbitrary value, in particular any value arbitrarily close to zero. Take any infinite product that converges to some value $alpha$. To let it converge to arbitrary $betainmathbb C$ instead, multiply the first factor by $beta/alpha$. In terms of your factors $1+a_n(s)$, you need
$$
1+a_n'(s)=fracbetaalpha(1+a_n(s))
$$
and thus
$$
a_n'(s)=fracbetaalpha(1+a_n(s))-1;.
$$
Edit in response to the new question:
The product can still be zero if $b_1=-1$ and $m_1=1$. Excluding this trivial case, whether the product can converge to zero now depends on $s$.
Since the product converges, $a_n$ converges to $0$; it follows that $sgt0$. Then the product is minimal if $b_n=-1$ and $m_n=n$ for all $nge2$, yielding
$$
prod_nge2left(1-n^-sright);,
$$
with another factor of $2$ if we choose to include $m_1=1$ with $b_1=1$. We can obtain a lower bound as follows:
begineqnarray*
prod_nge2left(1-n^-sright)
&=&
expleft(sum_nge2logleft(1-n^-sright)right)\
<&
expleft(-sum_nge2n^-sright)\
&=&expleft(1-zeta(s)right);.
endeqnarray*
This holds for $sgt1$. The zeta function has a pole at $s=1$, and in this case the sum in the exponent diverges logarithmically as $ntoinfty$. Thus, for $sle1$ the product can converge to zero.
answered Jul 19 at 4:40
joriki
164k10180328
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The product can take an arbitrary value, in particular any value arbitrarily close to zero. Take any infinite product that converges to some value $alpha$. To let it converge to arbitrary $betainmathbb C$ instead, multiply the first factor by $beta/alpha$. In terms of your factors $1+a_n(s)$, you need
$$
1+a_n'(s)=fracbetaalpha(1+a_n(s))
$$
and thus
$$
a_n'(s)=fracbetaalpha(1+a_n(s))-1;.
$$
Edit in response to the new question:
The product can still be zero if $b_1=-1$ and $m_1=1$. Excluding this trivial case, whether the product can converge to zero now depends on $s$.
Since the product converges, $a_n$ converges to $0$; it follows that $sgt0$. Then the product is minimal if $b_n=-1$ and $m_n=n$ for all $nge2$, yielding
$$
prod_nge2left(1-n^-sright);,
$$
with another factor of $2$ if we choose to include $m_1=1$ with $b_1=1$. We can obtain a lower bound as follows:
begineqnarray*
prod_nge2left(1-n^-sright)
&=&
expleft(sum_nge2logleft(1-n^-sright)right)\
<&
expleft(-sum_nge2n^-sright)\
&=&expleft(1-zeta(s)right);.
endeqnarray*
This holds for $sgt1$. The zeta function has a pole at $s=1$, and in this case the sum in the exponent diverges logarithmically as $ntoinfty$. Thus, for $sle1$ the product can converge to zero.
answered Jul 19 at 4:40
joriki
164k10180328
The product can take an arbitrary value, in particular any value arbitrarily close to zero. Take any infinite product that converges to some value $alpha$. To let it converge to arbitrary $betainmathbb C$ instead, multiply the first factor by $beta/alpha$. In terms of your factors $1+a_n(s)$, you need
$$
1+a_n'(s)=fracbetaalpha(1+a_n(s))
$$
and thus
$$
a_n'(s)=fracbetaalpha(1+a_n(s))-1;.
$$
Edit in response to the new question:
The product can still be zero if $b_1=-1$ and $m_1=1$. Excluding this trivial case, whether the product can converge to zero now depends on $s$.
Since the product converges, $a_n$ converges to $0$; it follows that $sgt0$. Then the product is minimal if $b_n=-1$ and $m_n=n$ for all $nge2$, yielding
$$
prod_nge2left(1-n^-sright);,
$$
with another factor of $2$ if we choose to include $m_1=1$ with $b_1=1$. We can obtain a lower bound as follows:
begineqnarray*
prod_nge2left(1-n^-sright)
&=&
expleft(sum_nge2logleft(1-n^-sright)right)\
<&
expleft(-sum_nge2n^-sright)\
&=&expleft(1-zeta(s)right);.
endeqnarray*
This holds for $sgt1$. The zeta function has a pole at $s=1$, and in this case the sum in the exponent diverges logarithmically as $ntoinfty$. Thus, for $sle1$ the product can converge to zero.
answered Jul 19 at 4:40
joriki
164k10180328
edited Jul 19 at 10:41
answered Jul 19 at 4:40
joriki
164k10180328
answered Jul 19 at 4:40
joriki
164k10180328
answered Jul 19 at 4:40
joriki
164k10180328
164k10180328
add a comment |Â
add a comment |Â
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I get the impression that you probably have conditions on the $a_n(s)$ in mind that you didn't state. As written, the $1+$ is irrelevant; you could just as well shift the $a_n(s)$ and write $prod_na_n(s)$, and you could trivially make the product come out as any desired value simply by changing the first factor accordingly.
– joriki
Jul 19 at 4:21
Then only condition is $a_n(s) to 0$ as $n to infty$.
– Grown pains
Jul 19 at 4:27
That's not an additional condition; that follows from the convergence of the product, which is already a condition in the question.
– joriki
Jul 19 at 4:36
Okay. What if $a_n(s)$ is of the form $c_nm_n^-s$ where $|c_n| = 1$ and $m_n$ is a subsequence of all naturals?
– Grown pains
Jul 19 at 4:43
The naturals aren't a sequence, so they don't have subsequences.
– joriki
Jul 19 at 4:48