Does the term of a convergent infinite sum create a convergent infinite product with 1 + term?
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Let
$0<sum_{n=k}^infty f(n)<infty$ where $f(n)>0,nge k$.
Then is the following guaranteed to be true:
$$0<prod_{n=k}^infty(f(n)+1)<infty$$
For example:
$$f(n)=cfrac{1}{n^2}$$ $$sum_{n=k}^infty f(n)=sum_{n=k}^infty cfrac{1}{n^2}=cfrac{pi^2}{6}$$
$$prod_{n=k}^infty(f(n)+1)=prod_{n=k}^infty cfrac{n^2+1}{n^2}approx 3.676$$
If there is not a guarantee (perhaps shown by a counter-example), then why?
real-analysis
edited Nov 13 at 23:57
amWhy
191k27223437
asked Nov 13 at 20:49
user3213847
32
add a comment |
up vote
0
down vote
favorite
Let
$0<sum_{n=k}^infty f(n)<infty$ where $f(n)>0,nge k$.
Then is the following guaranteed to be true:
$$0<prod_{n=k}^infty(f(n)+1)<infty$$
For example:
$$f(n)=cfrac{1}{n^2}$$ $$sum_{n=k}^infty f(n)=sum_{n=k}^infty cfrac{1}{n^2}=cfrac{pi^2}{6}$$
$$prod_{n=k}^infty(f(n)+1)=prod_{n=k}^infty cfrac{n^2+1}{n^2}approx 3.676$$
If there is not a guarantee (perhaps shown by a counter-example), then why?
real-analysis
edited Nov 13 at 23:57
amWhy
191k27223437
asked Nov 13 at 20:49
user3213847
32
Use the fact that $log(x+1) le x$ whenever $x ge 0$.
– Umberto P.
Nov 13 at 20:55
1
@UmbertoP. Elaborate on how that applies?
– user3213847
Nov 13 at 20:59
@user3213847 $ln left(prod_i a_iright) = sum_i ln a_i$.
– eyeballfrog
Nov 13 at 21:01
@user8268 one of the conditions is that $f(n)>0$
– user3213847
Nov 13 at 21:04
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let
$0<sum_{n=k}^infty f(n)<infty$ where $f(n)>0,nge k$.
Then is the following guaranteed to be true:
$$0<prod_{n=k}^infty(f(n)+1)<infty$$
For example:
$$f(n)=cfrac{1}{n^2}$$ $$sum_{n=k}^infty f(n)=sum_{n=k}^infty cfrac{1}{n^2}=cfrac{pi^2}{6}$$
$$prod_{n=k}^infty(f(n)+1)=prod_{n=k}^infty cfrac{n^2+1}{n^2}approx 3.676$$
If there is not a guarantee (perhaps shown by a counter-example), then why?
real-analysis
edited Nov 13 at 23:57
amWhy
191k27223437
asked Nov 13 at 20:49
user3213847
32
Let
$0<sum_{n=k}^infty f(n)<infty$ where $f(n)>0,nge k$.
Then is the following guaranteed to be true:
$$0<prod_{n=k}^infty(f(n)+1)<infty$$
For example:
$$f(n)=cfrac{1}{n^2}$$ $$sum_{n=k}^infty f(n)=sum_{n=k}^infty cfrac{1}{n^2}=cfrac{pi^2}{6}$$
$$prod_{n=k}^infty(f(n)+1)=prod_{n=k}^infty cfrac{n^2+1}{n^2}approx 3.676$$
If there is not a guarantee (perhaps shown by a counter-example), then why?
real-analysis
real-analysis
edited Nov 13 at 23:57
amWhy
191k27223437
asked Nov 13 at 20:49
user3213847
32
edited Nov 13 at 23:57
amWhy
191k27223437
asked Nov 13 at 20:49
user3213847
32
edited Nov 13 at 23:57
amWhy
191k27223437
edited Nov 13 at 23:57
amWhy
191k27223437
edited Nov 13 at 23:57
amWhy
191k27223437
191k27223437
asked Nov 13 at 20:49
user3213847
32
asked Nov 13 at 20:49
user3213847
32
asked Nov 13 at 20:49
user3213847
32
32
Use the fact that $log(x+1) le x$ whenever $x ge 0$.
– Umberto P.
Nov 13 at 20:55
1
@UmbertoP. Elaborate on how that applies?
– user3213847
Nov 13 at 20:59
@user3213847 $ln left(prod_i a_iright) = sum_i ln a_i$.
– eyeballfrog
Nov 13 at 21:01
@user8268 one of the conditions is that $f(n)>0$
– user3213847
Nov 13 at 21:04
add a comment |
Use the fact that $log(x+1) le x$ whenever $x ge 0$.
– Umberto P.
Nov 13 at 20:55
1
@UmbertoP. Elaborate on how that applies?
– user3213847
Nov 13 at 20:59
@user3213847 $ln left(prod_i a_iright) = sum_i ln a_i$.
– eyeballfrog
Nov 13 at 21:01
@user8268 one of the conditions is that $f(n)>0$
– user3213847
Nov 13 at 21:04
Use the fact that $log(x+1) le x$ whenever $x ge 0$.
– Umberto P.
Nov 13 at 20:55
Use the fact that $log(x+1) le x$ whenever $x ge 0$.
– Umberto P.
Nov 13 at 20:55
1
1
@UmbertoP. Elaborate on how that applies?
– user3213847
Nov 13 at 20:59
@UmbertoP. Elaborate on how that applies?
– user3213847
Nov 13 at 20:59
@user3213847 $ln left(prod_i a_iright) = sum_i ln a_i$.
– eyeballfrog
Nov 13 at 21:01
@user3213847 $ln left(prod_i a_iright) = sum_i ln a_i$.
– eyeballfrog
Nov 13 at 21:01
@user8268 one of the conditions is that $f(n)>0$
– user3213847
Nov 13 at 21:04
@user8268 one of the conditions is that $f(n)>0$
– user3213847
Nov 13 at 21:04
add a comment |
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
Yes, it's true (and further the infinite product will be positive): suppose you have a positive convergent series $sum_n a_n$. In particular, $lim_{ntoinfty} a_n = 0$, and therefore
$$
ln(1+a_n) = a_n + o(a_n) tag{$dagger$}
$$
Then
$$
ln prod_{n=1}^N (1+a_n) = sum_{n=1}^N ln(1+a_n)
$$
but the series $sum_{n=1}^infty ln(1+a_n)$ converges by theorems of comparison for positive series, recalling $(dagger)$. So
$$
lim_{Ntoinfty}prod_{n=1}^N (1+a_n) = expsum_{n=1}^infty ln(1+a_n) inmathbb{R}_+,.
$$
answered Nov 13 at 21:07
Clement C.
48.6k33784
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, it's true (and further the infinite product will be positive): suppose you have a positive convergent series $sum_n a_n$. In particular, $lim_{ntoinfty} a_n = 0$, and therefore
$$
ln(1+a_n) = a_n + o(a_n) tag{$dagger$}
$$
Then
$$
ln prod_{n=1}^N (1+a_n) = sum_{n=1}^N ln(1+a_n)
$$
but the series $sum_{n=1}^infty ln(1+a_n)$ converges by theorems of comparison for positive series, recalling $(dagger)$. So
$$
lim_{Ntoinfty}prod_{n=1}^N (1+a_n) = expsum_{n=1}^infty ln(1+a_n) inmathbb{R}_+,.
$$
answered Nov 13 at 21:07
Clement C.
48.6k33784
add a comment |
up vote
1
down vote
accepted
Yes, it's true (and further the infinite product will be positive): suppose you have a positive convergent series $sum_n a_n$. In particular, $lim_{ntoinfty} a_n = 0$, and therefore
$$
ln(1+a_n) = a_n + o(a_n) tag{$dagger$}
$$
Then
$$
ln prod_{n=1}^N (1+a_n) = sum_{n=1}^N ln(1+a_n)
$$
but the series $sum_{n=1}^infty ln(1+a_n)$ converges by theorems of comparison for positive series, recalling $(dagger)$. So
$$
lim_{Ntoinfty}prod_{n=1}^N (1+a_n) = expsum_{n=1}^infty ln(1+a_n) inmathbb{R}_+,.
$$
answered Nov 13 at 21:07
Clement C.
48.6k33784
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, it's true (and further the infinite product will be positive): suppose you have a positive convergent series $sum_n a_n$. In particular, $lim_{ntoinfty} a_n = 0$, and therefore
$$
ln(1+a_n) = a_n + o(a_n) tag{$dagger$}
$$
Then
$$
ln prod_{n=1}^N (1+a_n) = sum_{n=1}^N ln(1+a_n)
$$
but the series $sum_{n=1}^infty ln(1+a_n)$ converges by theorems of comparison for positive series, recalling $(dagger)$. So
$$
lim_{Ntoinfty}prod_{n=1}^N (1+a_n) = expsum_{n=1}^infty ln(1+a_n) inmathbb{R}_+,.
$$
answered Nov 13 at 21:07
Clement C.
48.6k33784
Yes, it's true (and further the infinite product will be positive): suppose you have a positive convergent series $sum_n a_n$. In particular, $lim_{ntoinfty} a_n = 0$, and therefore
$$
ln(1+a_n) = a_n + o(a_n) tag{$dagger$}
$$
Then
$$
ln prod_{n=1}^N (1+a_n) = sum_{n=1}^N ln(1+a_n)
$$
but the series $sum_{n=1}^infty ln(1+a_n)$ converges by theorems of comparison for positive series, recalling $(dagger)$. So
$$
lim_{Ntoinfty}prod_{n=1}^N (1+a_n) = expsum_{n=1}^infty ln(1+a_n) inmathbb{R}_+,.
$$
answered Nov 13 at 21:07
Clement C.
48.6k33784
answered Nov 13 at 21:07
Clement C.
48.6k33784
answered Nov 13 at 21:07
Clement C.
48.6k33784
answered Nov 13 at 21:07
Clement C.
48.6k33784
48.6k33784
add a comment |
add a comment |
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Use the fact that $log(x+1) le x$ whenever $x ge 0$.
– Umberto P.
Nov 13 at 20:55
1
@UmbertoP. Elaborate on how that applies?
– user3213847
Nov 13 at 20:59
@user3213847 $ln left(prod_i a_iright) = sum_i ln a_i$.
– eyeballfrog
Nov 13 at 21:01
@user8268 one of the conditions is that $f(n)>0$
– user3213847
Nov 13 at 21:04