Find the value of the infinite product: $prodlimits_n=1^inftyleft(1+frac13^nright)$
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up vote
7
down vote
favorite
I am Anay, here is a problem I am stuck with:
$$x = prodlimits_n=1^infty left ( 1 + frac13^n right )$$
The task is to find the value of $x$. (obviously, we aren't supposed to have infinite products or sums, etc. in the answer)
This is what I have done:
We define the sequence $a_k$ as,
$$a_k = prodlimits_n=1^k left ( 1 + frac13^n right )$$
First, we put some bounds on $a_k$, as it is a increasing sequence, we already have the lower bound as $frac43$. Now to get the higher bound, we have the following inequality for all integers $x$ (easily proved through binomial expansion):
$$left(1 + frac1xright)^x < e$$
So,
$$(1+x) < e^frac1x$$
Using this inequality many times, we have, (using the formula for sum of a geometric progression)
$$a_k < e^2$$
Thus,
$$frac43leq a_k< e^2$$
Then, to prove that this sequence is converging we show that it has the Cauchy Property. This can be done as follows:
First, we have,
$$a_k = a_k-1 + fraca_k-13^k$$
So adding such equations for $a_1$, $a_2$ ..... $a_k$, we see that all terms cancel out and the following remains:
$$a_k = a_1 + sumlimits_i = 1^k-1fraca_i3^i+1$$
So, if $m < n$,
$$a_n - a_m = sumlimits_i=m^n-1 fraca_i3^i+1 < a_n-1left(frac3^n - 3^m3^m+ntimes2 right)$$
As $a_k$ is a increasing sequence, we have used $a_n-1 > a_n-2>....>a_m $. And then we use the formula for sum of a geometric progression to get the result. Now we can see that when $m$ and $n$ are large enough, we can have the RHS arbitrarily small as $a_n-1$ has a upper bound ($e^2$), thus the sequence has the Cauchy property and it is converging.
After this I thought may be the sequence converges to the bound which I established ($e^2$), but it is not so as I checked it through a computer program, it approaches around $1.56$, which is far below $e^2$. So, after this I try many other methods to find where the sequence converges to but I found no luck. Also, I couldn't find any results on Google, so I have come to your help. How do I solve this?
Thanks in advance.
sequences-and-series convergence exponentiation cauchy-sequences infinite-product
edited Aug 30 at 2:08
Robson
47320
asked Dec 27 '16 at 17:54
Anay Karnik
1437
 |Â
show 6 more comments
up vote
7
down vote
favorite
I am Anay, here is a problem I am stuck with:
$$x = prodlimits_n=1^infty left ( 1 + frac13^n right )$$
The task is to find the value of $x$. (obviously, we aren't supposed to have infinite products or sums, etc. in the answer)
This is what I have done:
We define the sequence $a_k$ as,
$$a_k = prodlimits_n=1^k left ( 1 + frac13^n right )$$
First, we put some bounds on $a_k$, as it is a increasing sequence, we already have the lower bound as $frac43$. Now to get the higher bound, we have the following inequality for all integers $x$ (easily proved through binomial expansion):
$$left(1 + frac1xright)^x < e$$
So,
$$(1+x) < e^frac1x$$
Using this inequality many times, we have, (using the formula for sum of a geometric progression)
$$a_k < e^2$$
Thus,
$$frac43leq a_k< e^2$$
Then, to prove that this sequence is converging we show that it has the Cauchy Property. This can be done as follows:
First, we have,
$$a_k = a_k-1 + fraca_k-13^k$$
So adding such equations for $a_1$, $a_2$ ..... $a_k$, we see that all terms cancel out and the following remains:
$$a_k = a_1 + sumlimits_i = 1^k-1fraca_i3^i+1$$
So, if $m < n$,
$$a_n - a_m = sumlimits_i=m^n-1 fraca_i3^i+1 < a_n-1left(frac3^n - 3^m3^m+ntimes2 right)$$
As $a_k$ is a increasing sequence, we have used $a_n-1 > a_n-2>....>a_m $. And then we use the formula for sum of a geometric progression to get the result. Now we can see that when $m$ and $n$ are large enough, we can have the RHS arbitrarily small as $a_n-1$ has a upper bound ($e^2$), thus the sequence has the Cauchy property and it is converging.
After this I thought may be the sequence converges to the bound which I established ($e^2$), but it is not so as I checked it through a computer program, it approaches around $1.56$, which is far below $e^2$. So, after this I try many other methods to find where the sequence converges to but I found no luck. Also, I couldn't find any results on Google, so I have come to your help. How do I solve this?
Thanks in advance.
sequences-and-series convergence exponentiation cauchy-sequences infinite-product
edited Aug 30 at 2:08
Robson
47320
asked Dec 27 '16 at 17:54
Anay Karnik
1437
1
Wolfram Alpha thinks the result is ≈1.564934018567011537938849106728835416569425919895035009496...
– MatheMagic
Dec 27 '16 at 18:06
1
See: en.wikipedia.org/wiki/Euler_function and en.wikipedia.org/wiki/Q-Pochhammer_symbol In summary: there are no simple closed form for this product in elementary functions. Try to approximate it instead.
– Winther
Dec 27 '16 at 18:13
1
$displaystyleprod_n=1^inftyleft(1+frac13^nright)=fracleft(-1;frac13right)_infty2$.
– MatheMagic
Dec 27 '16 at 18:20
5
See the link:math.stackexchange.com/questions/1924882/…
– MatheMagic
Dec 27 '16 at 18:24
1
A good approximation is $$prod_n=1^inftyleft(1+frac13^nright) approx e^frac12cdot 3^kprod_n=1^kleft(1+frac13^nright)$$ for any integer $k$. The higher $k$ the better the approximation. For example for $k=4$ it gives $frac91840 sqrt[162]e59049 simeq 1.56495$ while the exact answer is $simeq 1.56493$.
– Winther
Dec 27 '16 at 18:25
 |Â
show 6 more comments
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I am Anay, here is a problem I am stuck with:
$$x = prodlimits_n=1^infty left ( 1 + frac13^n right )$$
The task is to find the value of $x$. (obviously, we aren't supposed to have infinite products or sums, etc. in the answer)
This is what I have done:
We define the sequence $a_k$ as,
$$a_k = prodlimits_n=1^k left ( 1 + frac13^n right )$$
First, we put some bounds on $a_k$, as it is a increasing sequence, we already have the lower bound as $frac43$. Now to get the higher bound, we have the following inequality for all integers $x$ (easily proved through binomial expansion):
$$left(1 + frac1xright)^x < e$$
So,
$$(1+x) < e^frac1x$$
Using this inequality many times, we have, (using the formula for sum of a geometric progression)
$$a_k < e^2$$
Thus,
$$frac43leq a_k< e^2$$
Then, to prove that this sequence is converging we show that it has the Cauchy Property. This can be done as follows:
First, we have,
$$a_k = a_k-1 + fraca_k-13^k$$
So adding such equations for $a_1$, $a_2$ ..... $a_k$, we see that all terms cancel out and the following remains:
$$a_k = a_1 + sumlimits_i = 1^k-1fraca_i3^i+1$$
So, if $m < n$,
$$a_n - a_m = sumlimits_i=m^n-1 fraca_i3^i+1 < a_n-1left(frac3^n - 3^m3^m+ntimes2 right)$$
As $a_k$ is a increasing sequence, we have used $a_n-1 > a_n-2>....>a_m $. And then we use the formula for sum of a geometric progression to get the result. Now we can see that when $m$ and $n$ are large enough, we can have the RHS arbitrarily small as $a_n-1$ has a upper bound ($e^2$), thus the sequence has the Cauchy property and it is converging.
After this I thought may be the sequence converges to the bound which I established ($e^2$), but it is not so as I checked it through a computer program, it approaches around $1.56$, which is far below $e^2$. So, after this I try many other methods to find where the sequence converges to but I found no luck. Also, I couldn't find any results on Google, so I have come to your help. How do I solve this?
Thanks in advance.
sequences-and-series convergence exponentiation cauchy-sequences infinite-product
edited Aug 30 at 2:08
Robson
47320
asked Dec 27 '16 at 17:54
Anay Karnik
1437
I am Anay, here is a problem I am stuck with:
$$x = prodlimits_n=1^infty left ( 1 + frac13^n right )$$
The task is to find the value of $x$. (obviously, we aren't supposed to have infinite products or sums, etc. in the answer)
This is what I have done:
We define the sequence $a_k$ as,
$$a_k = prodlimits_n=1^k left ( 1 + frac13^n right )$$
First, we put some bounds on $a_k$, as it is a increasing sequence, we already have the lower bound as $frac43$. Now to get the higher bound, we have the following inequality for all integers $x$ (easily proved through binomial expansion):
$$left(1 + frac1xright)^x < e$$
So,
$$(1+x) < e^frac1x$$
Using this inequality many times, we have, (using the formula for sum of a geometric progression)
$$a_k < e^2$$
Thus,
$$frac43leq a_k< e^2$$
Then, to prove that this sequence is converging we show that it has the Cauchy Property. This can be done as follows:
First, we have,
$$a_k = a_k-1 + fraca_k-13^k$$
So adding such equations for $a_1$, $a_2$ ..... $a_k$, we see that all terms cancel out and the following remains:
$$a_k = a_1 + sumlimits_i = 1^k-1fraca_i3^i+1$$
So, if $m < n$,
$$a_n - a_m = sumlimits_i=m^n-1 fraca_i3^i+1 < a_n-1left(frac3^n - 3^m3^m+ntimes2 right)$$
As $a_k$ is a increasing sequence, we have used $a_n-1 > a_n-2>....>a_m $. And then we use the formula for sum of a geometric progression to get the result. Now we can see that when $m$ and $n$ are large enough, we can have the RHS arbitrarily small as $a_n-1$ has a upper bound ($e^2$), thus the sequence has the Cauchy property and it is converging.
After this I thought may be the sequence converges to the bound which I established ($e^2$), but it is not so as I checked it through a computer program, it approaches around $1.56$, which is far below $e^2$. So, after this I try many other methods to find where the sequence converges to but I found no luck. Also, I couldn't find any results on Google, so I have come to your help. How do I solve this?
Thanks in advance.
sequences-and-series convergence exponentiation cauchy-sequences infinite-product
sequences-and-series convergence exponentiation cauchy-sequences infinite-product
edited Aug 30 at 2:08
Robson
47320
asked Dec 27 '16 at 17:54
Anay Karnik
1437
edited Aug 30 at 2:08
Robson
47320
asked Dec 27 '16 at 17:54
Anay Karnik
1437
edited Aug 30 at 2:08
Robson
47320
edited Aug 30 at 2:08
Robson
47320
edited Aug 30 at 2:08
Robson
47320
47320
asked Dec 27 '16 at 17:54
Anay Karnik
1437
asked Dec 27 '16 at 17:54
Anay Karnik
1437
asked Dec 27 '16 at 17:54
Anay Karnik
1437
1437
1
Wolfram Alpha thinks the result is ≈1.564934018567011537938849106728835416569425919895035009496...
– MatheMagic
Dec 27 '16 at 18:06
1
See: en.wikipedia.org/wiki/Euler_function and en.wikipedia.org/wiki/Q-Pochhammer_symbol In summary: there are no simple closed form for this product in elementary functions. Try to approximate it instead.
– Winther
Dec 27 '16 at 18:13
1
$displaystyleprod_n=1^inftyleft(1+frac13^nright)=fracleft(-1;frac13right)_infty2$.
– MatheMagic
Dec 27 '16 at 18:20
5
See the link:math.stackexchange.com/questions/1924882/…
– MatheMagic
Dec 27 '16 at 18:24
1
A good approximation is $$prod_n=1^inftyleft(1+frac13^nright) approx e^frac12cdot 3^kprod_n=1^kleft(1+frac13^nright)$$ for any integer $k$. The higher $k$ the better the approximation. For example for $k=4$ it gives $frac91840 sqrt[162]e59049 simeq 1.56495$ while the exact answer is $simeq 1.56493$.
– Winther
Dec 27 '16 at 18:25
 |Â
show 6 more comments
1
Wolfram Alpha thinks the result is ≈1.564934018567011537938849106728835416569425919895035009496...
– MatheMagic
Dec 27 '16 at 18:06
1
See: en.wikipedia.org/wiki/Euler_function and en.wikipedia.org/wiki/Q-Pochhammer_symbol In summary: there are no simple closed form for this product in elementary functions. Try to approximate it instead.
– Winther
Dec 27 '16 at 18:13
1
$displaystyleprod_n=1^inftyleft(1+frac13^nright)=fracleft(-1;frac13right)_infty2$.
– MatheMagic
Dec 27 '16 at 18:20
5
See the link:math.stackexchange.com/questions/1924882/…
– MatheMagic
Dec 27 '16 at 18:24
1
A good approximation is $$prod_n=1^inftyleft(1+frac13^nright) approx e^frac12cdot 3^kprod_n=1^kleft(1+frac13^nright)$$ for any integer $k$. The higher $k$ the better the approximation. For example for $k=4$ it gives $frac91840 sqrt[162]e59049 simeq 1.56495$ while the exact answer is $simeq 1.56493$.
– Winther
Dec 27 '16 at 18:25
1
1
Wolfram Alpha thinks the result is ≈1.564934018567011537938849106728835416569425919895035009496...
– MatheMagic
Dec 27 '16 at 18:06
Wolfram Alpha thinks the result is ≈1.564934018567011537938849106728835416569425919895035009496...
– MatheMagic
Dec 27 '16 at 18:06
1
1
See: en.wikipedia.org/wiki/Euler_function and en.wikipedia.org/wiki/Q-Pochhammer_symbol In summary: there are no simple closed form for this product in elementary functions. Try to approximate it instead.
– Winther
Dec 27 '16 at 18:13
See: en.wikipedia.org/wiki/Euler_function and en.wikipedia.org/wiki/Q-Pochhammer_symbol In summary: there are no simple closed form for this product in elementary functions. Try to approximate it instead.
– Winther
Dec 27 '16 at 18:13
1
1
$displaystyleprod_n=1^inftyleft(1+frac13^nright)=fracleft(-1;frac13right)_infty2$.
– MatheMagic
Dec 27 '16 at 18:20
$displaystyleprod_n=1^inftyleft(1+frac13^nright)=fracleft(-1;frac13right)_infty2$.
– MatheMagic
Dec 27 '16 at 18:20
5
5
See the link:math.stackexchange.com/questions/1924882/…
– MatheMagic
Dec 27 '16 at 18:24
See the link:math.stackexchange.com/questions/1924882/…
– MatheMagic
Dec 27 '16 at 18:24
1
1
A good approximation is $$prod_n=1^inftyleft(1+frac13^nright) approx e^frac12cdot 3^kprod_n=1^kleft(1+frac13^nright)$$ for any integer $k$. The higher $k$ the better the approximation. For example for $k=4$ it gives $frac91840 sqrt[162]e59049 simeq 1.56495$ while the exact answer is $simeq 1.56493$.
– Winther
Dec 27 '16 at 18:25
A good approximation is $$prod_n=1^inftyleft(1+frac13^nright) approx e^frac12cdot 3^kprod_n=1^kleft(1+frac13^nright)$$ for any integer $k$. The higher $k$ the better the approximation. For example for $k=4$ it gives $frac91840 sqrt[162]e59049 simeq 1.56495$ while the exact answer is $simeq 1.56493$.
– Winther
Dec 27 '16 at 18:25
 |Â
show 6 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
$$left(1+dfrac13right)left(1+dfrac13^2right)=1+frac13+frac13^2+frac13^3$$
$$left(1+dfrac13right)left(1+dfrac13^2right)left(1+dfrac13^3right)=1+frac13+frac13^2+frac23^3+frac13^4+frac13^5+frac13^6$$
I guess continuing this you can observe a pattern and it will leads to the solution.
Good luck
EDIT:
Above pattern shows us, the infinite summation can be written as $$sum_n=1^inftydfracq(n)3^n,$$ where $q$ is the partition function. So there is no closed form involving elementary functions.
answered Dec 27 '16 at 18:21
Bumblebee
9,45512449
There is no simple pattern that allows a simple solution here. The solution can formally be written in terms of en.wikipedia.org/wiki/Q-Pochhammer_symbol or en.wikipedia.org/wiki/Euler_function - these are non-elementary functions (defined as infinite products) and do not have a simple representation.
– Winther
Dec 27 '16 at 18:28
Yes. Just now I realize that the pattern involves the partition function and it is non-elementary.
– Bumblebee
Dec 27 '16 at 18:31
uh? I don't see how this pattern can help in a infinite sequence
– Anay Karnik
Dec 27 '16 at 18:32
We define Infinite products as the limit of a finite product. So there is a relationship between this pattern and your infinite product. :)
– Bumblebee
Dec 27 '16 at 18:35
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
$$left(1+dfrac13right)left(1+dfrac13^2right)=1+frac13+frac13^2+frac13^3$$
$$left(1+dfrac13right)left(1+dfrac13^2right)left(1+dfrac13^3right)=1+frac13+frac13^2+frac23^3+frac13^4+frac13^5+frac13^6$$
I guess continuing this you can observe a pattern and it will leads to the solution.
Good luck
EDIT:
Above pattern shows us, the infinite summation can be written as $$sum_n=1^inftydfracq(n)3^n,$$ where $q$ is the partition function. So there is no closed form involving elementary functions.
answered Dec 27 '16 at 18:21
Bumblebee
9,45512449
There is no simple pattern that allows a simple solution here. The solution can formally be written in terms of en.wikipedia.org/wiki/Q-Pochhammer_symbol or en.wikipedia.org/wiki/Euler_function - these are non-elementary functions (defined as infinite products) and do not have a simple representation.
– Winther
Dec 27 '16 at 18:28
Yes. Just now I realize that the pattern involves the partition function and it is non-elementary.
– Bumblebee
Dec 27 '16 at 18:31
uh? I don't see how this pattern can help in a infinite sequence
– Anay Karnik
Dec 27 '16 at 18:32
We define Infinite products as the limit of a finite product. So there is a relationship between this pattern and your infinite product. :)
– Bumblebee
Dec 27 '16 at 18:35
add a comment |Â
up vote
1
down vote
$$left(1+dfrac13right)left(1+dfrac13^2right)=1+frac13+frac13^2+frac13^3$$
$$left(1+dfrac13right)left(1+dfrac13^2right)left(1+dfrac13^3right)=1+frac13+frac13^2+frac23^3+frac13^4+frac13^5+frac13^6$$
I guess continuing this you can observe a pattern and it will leads to the solution.
Good luck
EDIT:
Above pattern shows us, the infinite summation can be written as $$sum_n=1^inftydfracq(n)3^n,$$ where $q$ is the partition function. So there is no closed form involving elementary functions.
answered Dec 27 '16 at 18:21
Bumblebee
9,45512449
There is no simple pattern that allows a simple solution here. The solution can formally be written in terms of en.wikipedia.org/wiki/Q-Pochhammer_symbol or en.wikipedia.org/wiki/Euler_function - these are non-elementary functions (defined as infinite products) and do not have a simple representation.
– Winther
Dec 27 '16 at 18:28
Yes. Just now I realize that the pattern involves the partition function and it is non-elementary.
– Bumblebee
Dec 27 '16 at 18:31
uh? I don't see how this pattern can help in a infinite sequence
– Anay Karnik
Dec 27 '16 at 18:32
We define Infinite products as the limit of a finite product. So there is a relationship between this pattern and your infinite product. :)
– Bumblebee
Dec 27 '16 at 18:35
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$left(1+dfrac13right)left(1+dfrac13^2right)=1+frac13+frac13^2+frac13^3$$
$$left(1+dfrac13right)left(1+dfrac13^2right)left(1+dfrac13^3right)=1+frac13+frac13^2+frac23^3+frac13^4+frac13^5+frac13^6$$
I guess continuing this you can observe a pattern and it will leads to the solution.
Good luck
EDIT:
Above pattern shows us, the infinite summation can be written as $$sum_n=1^inftydfracq(n)3^n,$$ where $q$ is the partition function. So there is no closed form involving elementary functions.
answered Dec 27 '16 at 18:21
Bumblebee
9,45512449
$$left(1+dfrac13right)left(1+dfrac13^2right)=1+frac13+frac13^2+frac13^3$$
$$left(1+dfrac13right)left(1+dfrac13^2right)left(1+dfrac13^3right)=1+frac13+frac13^2+frac23^3+frac13^4+frac13^5+frac13^6$$
I guess continuing this you can observe a pattern and it will leads to the solution.
Good luck
EDIT:
Above pattern shows us, the infinite summation can be written as $$sum_n=1^inftydfracq(n)3^n,$$ where $q$ is the partition function. So there is no closed form involving elementary functions.
answered Dec 27 '16 at 18:21
Bumblebee
9,45512449
edited Dec 27 '16 at 18:40
answered Dec 27 '16 at 18:21
Bumblebee
9,45512449
answered Dec 27 '16 at 18:21
Bumblebee
9,45512449
answered Dec 27 '16 at 18:21
Bumblebee
9,45512449
9,45512449
There is no simple pattern that allows a simple solution here. The solution can formally be written in terms of en.wikipedia.org/wiki/Q-Pochhammer_symbol or en.wikipedia.org/wiki/Euler_function - these are non-elementary functions (defined as infinite products) and do not have a simple representation.
– Winther
Dec 27 '16 at 18:28
Yes. Just now I realize that the pattern involves the partition function and it is non-elementary.
– Bumblebee
Dec 27 '16 at 18:31
uh? I don't see how this pattern can help in a infinite sequence
– Anay Karnik
Dec 27 '16 at 18:32
We define Infinite products as the limit of a finite product. So there is a relationship between this pattern and your infinite product. :)
– Bumblebee
Dec 27 '16 at 18:35
add a comment |Â
There is no simple pattern that allows a simple solution here. The solution can formally be written in terms of en.wikipedia.org/wiki/Q-Pochhammer_symbol or en.wikipedia.org/wiki/Euler_function - these are non-elementary functions (defined as infinite products) and do not have a simple representation.
– Winther
Dec 27 '16 at 18:28
Yes. Just now I realize that the pattern involves the partition function and it is non-elementary.
– Bumblebee
Dec 27 '16 at 18:31
uh? I don't see how this pattern can help in a infinite sequence
– Anay Karnik
Dec 27 '16 at 18:32
We define Infinite products as the limit of a finite product. So there is a relationship between this pattern and your infinite product. :)
– Bumblebee
Dec 27 '16 at 18:35
There is no simple pattern that allows a simple solution here. The solution can formally be written in terms of en.wikipedia.org/wiki/Q-Pochhammer_symbol or en.wikipedia.org/wiki/Euler_function - these are non-elementary functions (defined as infinite products) and do not have a simple representation.
– Winther
Dec 27 '16 at 18:28
There is no simple pattern that allows a simple solution here. The solution can formally be written in terms of en.wikipedia.org/wiki/Q-Pochhammer_symbol or en.wikipedia.org/wiki/Euler_function - these are non-elementary functions (defined as infinite products) and do not have a simple representation.
– Winther
Dec 27 '16 at 18:28
Yes. Just now I realize that the pattern involves the partition function and it is non-elementary.
– Bumblebee
Dec 27 '16 at 18:31
Yes. Just now I realize that the pattern involves the partition function and it is non-elementary.
– Bumblebee
Dec 27 '16 at 18:31
uh? I don't see how this pattern can help in a infinite sequence
– Anay Karnik
Dec 27 '16 at 18:32
uh? I don't see how this pattern can help in a infinite sequence
– Anay Karnik
Dec 27 '16 at 18:32
We define Infinite products as the limit of a finite product. So there is a relationship between this pattern and your infinite product. :)
– Bumblebee
Dec 27 '16 at 18:35
We define Infinite products as the limit of a finite product. So there is a relationship between this pattern and your infinite product. :)
– Bumblebee
Dec 27 '16 at 18:35
add a comment |Â
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1
Wolfram Alpha thinks the result is ≈1.564934018567011537938849106728835416569425919895035009496...
– MatheMagic
Dec 27 '16 at 18:06
1
See: en.wikipedia.org/wiki/Euler_function and en.wikipedia.org/wiki/Q-Pochhammer_symbol In summary: there are no simple closed form for this product in elementary functions. Try to approximate it instead.
– Winther
Dec 27 '16 at 18:13
1
$displaystyleprod_n=1^inftyleft(1+frac13^nright)=fracleft(-1;frac13right)_infty2$.
– MatheMagic
Dec 27 '16 at 18:20
5
See the link:math.stackexchange.com/questions/1924882/…
– MatheMagic
Dec 27 '16 at 18:24
1
A good approximation is $$prod_n=1^inftyleft(1+frac13^nright) approx e^frac12cdot 3^kprod_n=1^kleft(1+frac13^nright)$$ for any integer $k$. The higher $k$ the better the approximation. For example for $k=4$ it gives $frac91840 sqrt[162]e59049 simeq 1.56495$ while the exact answer is $simeq 1.56493$.
– Winther
Dec 27 '16 at 18:25