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Does the infinite product $frac21timesfrac43timesfrac65timesdots$ converge?
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I was wondering whether the infinite product
$$prod_i=0^inftyfrac2n + 22n + 1 = frac21timesfrac43timesfrac65timesdots$$
converges. For all I know it is certainly not absolutely convergent, but is it at least conditionally so? Telescoping is not possible in this case. I know that
$$1 - frac12 + frac13 - frac14 + dots = ln2$$
but if I take the logarithm off this infinite product, I'd get
$$ ln1 - lnfrac12 + lnfrac13 - lnfrac14 + dots$$
which is not conclusive.
convergence infinite-product
asked Jul 16 at 6:48
Voile
1132
 |Â
show 3 more comments
up vote
2
down vote
favorite
1
I was wondering whether the infinite product
$$prod_i=0^inftyfrac2n + 22n + 1 = frac21timesfrac43timesfrac65timesdots$$
converges. For all I know it is certainly not absolutely convergent, but is it at least conditionally so? Telescoping is not possible in this case. I know that
$$1 - frac12 + frac13 - frac14 + dots = ln2$$
but if I take the logarithm off this infinite product, I'd get
$$ ln1 - lnfrac12 + lnfrac13 - lnfrac14 + dots$$
which is not conclusive.
convergence infinite-product
asked Jul 16 at 6:48
Voile
1132
1
All the terms are greater than $1$, so there is no difference between conditional and absolute convergence. Maybe you had something else in mind?
– Eric Wofsey
Jul 16 at 6:52
1
Not a duplicate, but here is another question that use the same expression: math.stackexchange.com/questions/855990/…
– Trần Thúc Minh TrÃ
Jul 16 at 6:55
@EricWofsey: It doesn't answer the question: I can arrange it as $lnfrac21 + lnfrac43 + lnfrac65 + dots$ where the terms approachs 0, and I think ratio test gives 1.
– Voile
Jul 16 at 6:55
2
Probably not much of a help, but it's basically $lim_nto infty frac(2^n n!)^2(2n)!$.
– Sil
Jul 16 at 7:07
1
@Voile But it does answer the question... Since all terms are strictly greater than $1$, if it is conditionally convergent, it must also be absolutely convergent. Since you say that it is not absolutely convergent, it follows that it is not conditionally convergent. The product diverges.
– Eff
Jul 16 at 7:11
 |Â
show 3 more comments
up vote
2
down vote
favorite
1
up vote
2
down vote
favorite
1
1
I was wondering whether the infinite product
$$prod_i=0^inftyfrac2n + 22n + 1 = frac21timesfrac43timesfrac65timesdots$$
converges. For all I know it is certainly not absolutely convergent, but is it at least conditionally so? Telescoping is not possible in this case. I know that
$$1 - frac12 + frac13 - frac14 + dots = ln2$$
but if I take the logarithm off this infinite product, I'd get
$$ ln1 - lnfrac12 + lnfrac13 - lnfrac14 + dots$$
which is not conclusive.
convergence infinite-product
asked Jul 16 at 6:48
Voile
1132
I was wondering whether the infinite product
$$prod_i=0^inftyfrac2n + 22n + 1 = frac21timesfrac43timesfrac65timesdots$$
converges. For all I know it is certainly not absolutely convergent, but is it at least conditionally so? Telescoping is not possible in this case. I know that
$$1 - frac12 + frac13 - frac14 + dots = ln2$$
but if I take the logarithm off this infinite product, I'd get
$$ ln1 - lnfrac12 + lnfrac13 - lnfrac14 + dots$$
which is not conclusive.
convergence infinite-product
asked Jul 16 at 6:48
Voile
1132
asked Jul 16 at 6:48
Voile
1132
asked Jul 16 at 6:48
Voile
1132
asked Jul 16 at 6:48
Voile
1132
1132
1
All the terms are greater than $1$, so there is no difference between conditional and absolute convergence. Maybe you had something else in mind?
– Eric Wofsey
Jul 16 at 6:52
1
Not a duplicate, but here is another question that use the same expression: math.stackexchange.com/questions/855990/…
– Trần Thúc Minh TrÃ
Jul 16 at 6:55
@EricWofsey: It doesn't answer the question: I can arrange it as $lnfrac21 + lnfrac43 + lnfrac65 + dots$ where the terms approachs 0, and I think ratio test gives 1.
– Voile
Jul 16 at 6:55
2
Probably not much of a help, but it's basically $lim_nto infty frac(2^n n!)^2(2n)!$.
– Sil
Jul 16 at 7:07
1
@Voile But it does answer the question... Since all terms are strictly greater than $1$, if it is conditionally convergent, it must also be absolutely convergent. Since you say that it is not absolutely convergent, it follows that it is not conditionally convergent. The product diverges.
– Eff
Jul 16 at 7:11
 |Â
show 3 more comments
1
All the terms are greater than $1$, so there is no difference between conditional and absolute convergence. Maybe you had something else in mind?
– Eric Wofsey
Jul 16 at 6:52
1
Not a duplicate, but here is another question that use the same expression: math.stackexchange.com/questions/855990/…
– Trần Thúc Minh TrÃ
Jul 16 at 6:55
@EricWofsey: It doesn't answer the question: I can arrange it as $lnfrac21 + lnfrac43 + lnfrac65 + dots$ where the terms approachs 0, and I think ratio test gives 1.
– Voile
Jul 16 at 6:55
2
Probably not much of a help, but it's basically $lim_nto infty frac(2^n n!)^2(2n)!$.
– Sil
Jul 16 at 7:07
1
@Voile But it does answer the question... Since all terms are strictly greater than $1$, if it is conditionally convergent, it must also be absolutely convergent. Since you say that it is not absolutely convergent, it follows that it is not conditionally convergent. The product diverges.
– Eff
Jul 16 at 7:11
1
1
All the terms are greater than $1$, so there is no difference between conditional and absolute convergence. Maybe you had something else in mind?
– Eric Wofsey
Jul 16 at 6:52
All the terms are greater than $1$, so there is no difference between conditional and absolute convergence. Maybe you had something else in mind?
– Eric Wofsey
Jul 16 at 6:52
1
1
Not a duplicate, but here is another question that use the same expression: math.stackexchange.com/questions/855990/…
– Trần Thúc Minh TrÃ
Jul 16 at 6:55
Not a duplicate, but here is another question that use the same expression: math.stackexchange.com/questions/855990/…
– Trần Thúc Minh TrÃ
Jul 16 at 6:55
@EricWofsey: It doesn't answer the question: I can arrange it as $lnfrac21 + lnfrac43 + lnfrac65 + dots$ where the terms approachs 0, and I think ratio test gives 1.
– Voile
Jul 16 at 6:55
@EricWofsey: It doesn't answer the question: I can arrange it as $lnfrac21 + lnfrac43 + lnfrac65 + dots$ where the terms approachs 0, and I think ratio test gives 1.
– Voile
Jul 16 at 6:55
2
2
Probably not much of a help, but it's basically $lim_nto infty frac(2^n n!)^2(2n)!$.
– Sil
Jul 16 at 7:07
Probably not much of a help, but it's basically $lim_nto infty frac(2^n n!)^2(2n)!$.
– Sil
Jul 16 at 7:07
1
1
@Voile But it does answer the question... Since all terms are strictly greater than $1$, if it is conditionally convergent, it must also be absolutely convergent. Since you say that it is not absolutely convergent, it follows that it is not conditionally convergent. The product diverges.
– Eff
Jul 16 at 7:11
@Voile But it does answer the question... Since all terms are strictly greater than $1$, if it is conditionally convergent, it must also be absolutely convergent. Since you say that it is not absolutely convergent, it follows that it is not conditionally convergent. The product diverges.
– Eff
Jul 16 at 7:11
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
9
down vote
accepted
Hint: If $a,b,c,ldots$ are all positive
$$(1+a)(1+b)(1+c)cdotsge 1+a+b+c+cdots.$$
answered Jul 16 at 6:57
Jyrki Lahtonen
105k12161355
2
The general theme at play here is to always compare the convergence of $prod_i(1+a_i)$ and $sum_ia_i$. Surprisingly I didn't find this covered explicitly in our question abouot infinite products.
– Jyrki Lahtonen
Jul 16 at 7:15
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
Hint: If $a,b,c,ldots$ are all positive
$$(1+a)(1+b)(1+c)cdotsge 1+a+b+c+cdots.$$
answered Jul 16 at 6:57
Jyrki Lahtonen
105k12161355
2
The general theme at play here is to always compare the convergence of $prod_i(1+a_i)$ and $sum_ia_i$. Surprisingly I didn't find this covered explicitly in our question abouot infinite products.
– Jyrki Lahtonen
Jul 16 at 7:15
add a comment |Â
up vote
9
down vote
accepted
Hint: If $a,b,c,ldots$ are all positive
$$(1+a)(1+b)(1+c)cdotsge 1+a+b+c+cdots.$$
answered Jul 16 at 6:57
Jyrki Lahtonen
105k12161355
2
The general theme at play here is to always compare the convergence of $prod_i(1+a_i)$ and $sum_ia_i$. Surprisingly I didn't find this covered explicitly in our question abouot infinite products.
– Jyrki Lahtonen
Jul 16 at 7:15
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
Hint: If $a,b,c,ldots$ are all positive
$$(1+a)(1+b)(1+c)cdotsge 1+a+b+c+cdots.$$
answered Jul 16 at 6:57
Jyrki Lahtonen
105k12161355
Hint: If $a,b,c,ldots$ are all positive
$$(1+a)(1+b)(1+c)cdotsge 1+a+b+c+cdots.$$
answered Jul 16 at 6:57
Jyrki Lahtonen
105k12161355
answered Jul 16 at 6:57
Jyrki Lahtonen
105k12161355
answered Jul 16 at 6:57
Jyrki Lahtonen
105k12161355
answered Jul 16 at 6:57
Jyrki Lahtonen
105k12161355
105k12161355
2
The general theme at play here is to always compare the convergence of $prod_i(1+a_i)$ and $sum_ia_i$. Surprisingly I didn't find this covered explicitly in our question abouot infinite products.
– Jyrki Lahtonen
Jul 16 at 7:15
add a comment |Â
2
The general theme at play here is to always compare the convergence of $prod_i(1+a_i)$ and $sum_ia_i$. Surprisingly I didn't find this covered explicitly in our question abouot infinite products.
– Jyrki Lahtonen
Jul 16 at 7:15
2
2
The general theme at play here is to always compare the convergence of $prod_i(1+a_i)$ and $sum_ia_i$. Surprisingly I didn't find this covered explicitly in our question abouot infinite products.
– Jyrki Lahtonen
Jul 16 at 7:15
The general theme at play here is to always compare the convergence of $prod_i(1+a_i)$ and $sum_ia_i$. Surprisingly I didn't find this covered explicitly in our question abouot infinite products.
– Jyrki Lahtonen
Jul 16 at 7:15
add a comment |Â
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1
All the terms are greater than $1$, so there is no difference between conditional and absolute convergence. Maybe you had something else in mind?
– Eric Wofsey
Jul 16 at 6:52
1
Not a duplicate, but here is another question that use the same expression: math.stackexchange.com/questions/855990/…
– Trần Thúc Minh TrÃ
Jul 16 at 6:55
@EricWofsey: It doesn't answer the question: I can arrange it as $lnfrac21 + lnfrac43 + lnfrac65 + dots$ where the terms approachs 0, and I think ratio test gives 1.
– Voile
Jul 16 at 6:55
2
Probably not much of a help, but it's basically $lim_nto infty frac(2^n n!)^2(2n)!$.
– Sil
Jul 16 at 7:07
1
@Voile But it does answer the question... Since all terms are strictly greater than $1$, if it is conditionally convergent, it must also be absolutely convergent. Since you say that it is not absolutely convergent, it follows that it is not conditionally convergent. The product diverges.
– Eff
Jul 16 at 7:11