closed form of $prod_n=1^inftyleft(fracnn+1right)^(-1)^n-1n$Evaluating a trigonometric product $prod_n=1^inftycos^2left(frac1n^2right)$Closed form for the infinite product $prodlimits_k=0^infty left( 1-x^2^k right)$Closed form of infinite product $prodlimits_k=0^infty 2 left(1-fracx^1/2^k+11+x^1/2^k right)$Does $prod_t=1^inftyleft(1-frac11.127^tright)$ converge to a non-zero value?Closed form for $prod_i=2^infty (1 - frac1i!)$How to compute $prod_n=2^infty left(1-frac1n^nright)$?Closed form for the series $sum_k=1^infty (-1)^k ln left( tanh fracpi k x2 right)$Find closed form of Wallis's product type $prod_n=1^inftyleft(prod_k=0^m(n+k)^mchoose k(-1)^kright)^(-1)^n$Finding the closed form of the product $prod_i = 1^infty (1 - 1/q^i) $.Is the product $prod_n=1^inftyleft(fracGamma(n+frac12)sqrtnGamma(n)right)$ convergent?
Do I have to take mana from my deck or hand when tapping a dual land?
Anime with legendary swords made from talismans and a man who could change them with a shattered body
Why is the principal energy of an electron lower for excited electrons in a higher energy state?
ContourPlot — How do I color by contour curvature?
Should I assume I have passed probation?
Showing mass murder in a kid's book
Why is participating in the European Parliamentary elections used as a threat?
What does "tick" mean in this sentence?
Why do Radio Buttons not fill the entire outer circle?
Why would five hundred and five be same as one?
Did I make a mistake by ccing email to boss to others?
"Oh no!" in Latin
How much do grades matter for a future academia position?
Air travel with refrigerated insulin
How to reduce predictors the right way for a logistic regression model
Alignment of six matrices
Unable to disable Microsoft Store in domain environment
What the heck is gets(stdin) on site coderbyte?
Pre-Employment Background Check With Consent For Future Checks
Why can't the Brexit deadlock in the UK parliament be solved with a plurality vote?
Echo with obfuscation
Why the "ls" command is showing the permissions of files in a FAT32 partition?
How to get directions in deep space?
Do you waste sorcery points if you try to apply metamagic to a spell from a scroll but fail to cast it?
closed form of $prod_n=1^inftyleft(fracnn+1right)^(-1)^n-1n$
Evaluating a trigonometric product $prod_n=1^inftycos^2left(frac1n^2right)$Closed form for the infinite product $prodlimits_k=0^infty left( 1-x^2^k right)$Closed form of infinite product $prodlimits_k=0^infty 2 left(1-fracx^1/2^k+11+x^1/2^k right)$Does $prod_t=1^inftyleft(1-frac11.127^tright)$ converge to a non-zero value?Closed form for $prod_i=2^infty (1 - frac1i!)$How to compute $prod_n=2^infty left(1-frac1n^nright)$?Closed form for the series $sum_k=1^infty (-1)^k ln left( tanh fracpi k x2 right)$Find closed form of Wallis's product type $prod_n=1^inftyleft(prod_k=0^m(n+k)^mchoose k(-1)^kright)^(-1)^n$Finding the closed form of the product $prod_i = 1^infty (1 - 1/q^i) $.Is the product $prod_n=1^inftyleft(fracGamma(n+frac12)sqrtnGamma(n)right)$ convergent?
$begingroup$
I am looking for the closed form of this product.
$$prod_n=1^inftyleft(fracnn+1right)^(-1)^n-1n$$
I have sees it somewhere before but I can't remember it closed form. I remember the Glaisher's constant it is invloved alone with $2^7/6$ and maybe e (exponential function constant) also.
Does anyone knows it closed form?
infinite-product
asked Mar 14 at 6:56
coffeeecoffeee
14518
$endgroup$
add a comment |
$begingroup$
I am looking for the closed form of this product.
$$prod_n=1^inftyleft(fracnn+1right)^(-1)^n-1n$$
I have sees it somewhere before but I can't remember it closed form. I remember the Glaisher's constant it is invloved alone with $2^7/6$ and maybe e (exponential function constant) also.
Does anyone knows it closed form?
infinite-product
asked Mar 14 at 6:56
coffeeecoffeee
14518
$endgroup$
$begingroup$
Is it convergent? I got the product to be $$frac 3^5 cdot 5^9 cdot 7^13 cdots 2^3 cdot 4^7 cdot 6^11 cdots$$ which is definitely divergent.
$endgroup$
– Dbchatto67
Mar 14 at 7:32
$begingroup$
There are two limits possible, please see my answer.
$endgroup$
– user90369
Mar 14 at 16:51
add a comment |
$begingroup$
I am looking for the closed form of this product.
$$prod_n=1^inftyleft(fracnn+1right)^(-1)^n-1n$$
I have sees it somewhere before but I can't remember it closed form. I remember the Glaisher's constant it is invloved alone with $2^7/6$ and maybe e (exponential function constant) also.
Does anyone knows it closed form?
infinite-product
asked Mar 14 at 6:56
coffeeecoffeee
14518
$endgroup$
I am looking for the closed form of this product.
$$prod_n=1^inftyleft(fracnn+1right)^(-1)^n-1n$$
I have sees it somewhere before but I can't remember it closed form. I remember the Glaisher's constant it is invloved alone with $2^7/6$ and maybe e (exponential function constant) also.
Does anyone knows it closed form?
infinite-product
infinite-product
asked Mar 14 at 6:56
coffeeecoffeee
14518
asked Mar 14 at 6:56
coffeeecoffeee
14518
asked Mar 14 at 6:56
coffeeecoffeee
14518
asked Mar 14 at 6:56
coffeeecoffeee
14518
asked Mar 14 at 6:56
coffeeecoffeee
14518
14518
$begingroup$
Is it convergent? I got the product to be $$frac 3^5 cdot 5^9 cdot 7^13 cdots 2^3 cdot 4^7 cdot 6^11 cdots$$ which is definitely divergent.
$endgroup$
– Dbchatto67
Mar 14 at 7:32
$begingroup$
There are two limits possible, please see my answer.
$endgroup$
– user90369
Mar 14 at 16:51
add a comment |
$begingroup$
Is it convergent? I got the product to be $$frac 3^5 cdot 5^9 cdot 7^13 cdots 2^3 cdot 4^7 cdot 6^11 cdots$$ which is definitely divergent.
$endgroup$
– Dbchatto67
Mar 14 at 7:32
$begingroup$
There are two limits possible, please see my answer.
$endgroup$
– user90369
Mar 14 at 16:51
$begingroup$
Is it convergent? I got the product to be $$frac 3^5 cdot 5^9 cdot 7^13 cdots 2^3 cdot 4^7 cdot 6^11 cdots$$ which is definitely divergent.
$endgroup$
– Dbchatto67
Mar 14 at 7:32
$begingroup$
Is it convergent? I got the product to be $$frac 3^5 cdot 5^9 cdot 7^13 cdots 2^3 cdot 4^7 cdot 6^11 cdots$$ which is definitely divergent.
$endgroup$
– Dbchatto67
Mar 14 at 7:32
$begingroup$
There are two limits possible, please see my answer.
$endgroup$
– user90369
Mar 14 at 16:51
$begingroup$
There are two limits possible, please see my answer.
$endgroup$
– user90369
Mar 14 at 16:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$displaystyle 1/prodlimits_n=1^2Nleft(fracnn+1right)^(-1)^n-1n = frac1sqrt2 left( frace^N/2N^-1/8 prodlimits_n=1^Nleft(1+frac12nright)^n
right)^4 left( frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n right)^-1 $
$displaystyle limlimits_Ntoinfty frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n = limlimits_Ntoinftyfrace^N N^-1/2 prodlimits_n=1^Nleft(1+frac1nright)^n = fracsqrt2pieenspaceenspace$ (e.g. by the Stirling formula)
The first calculation formula of Glaisher for the constant named after him (but written here more compact with products instead of series) is:
$$A=2^1/36pi^1/6left( limlimits_Ntoinfty frace^N/2N^-1/8 prodlimits_n=1^Nleft(1+frac12nright)^n
right)^2/3left( limlimits_Ntoinfty frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n right)^-1/3$$
(see Glaisher page 46 formula (7))
We potentiate this equation of Glaisher on both sides with 6 and multiply one time the left side with $fracsqrt2pie$ and the right with it’s product. After a few simple elementary conversions follows:
$displaystyle limlimits_Ntoinftyprodlimits_n=1^2Nleft(fracnn+1right)^(-1)^n-1n = 2^1/6pi^1/2eA^-6 approx 1.2157517513…$
$displaystyle limlimits_Ntoinftyprodlimits_n=1^2N+1left(fracnn+1right)^(-1)^n-1n = 2^1/6pi^1/2A^-6 approx 0.44725…$
answered Mar 14 at 16:08
user90369user90369
8,475925
$endgroup$
$begingroup$
thank you for showing the full answer
$endgroup$
– coffeee
Mar 15 at 20:31
$begingroup$
@coffeee : You are welcome, it was a pleasure. ;)
$endgroup$
– user90369
Mar 15 at 22:26
add a comment |
$begingroup$
Let $$a_n=left(fracnn+1right)^(-1)^n-1 n$$ then
$$a_2p= left(frac2p2 p+1right)^-2 pqquad textandqquad a_2p+1=left(frac2 p+12 p+2right)^2 p+1$$
Now, using a CAS,
$$prod_p=1^m a_2p=fracsqrt[12]2 sqrtpi exp left(-2 zeta ^(1,0)(-1,m+1)+2 zeta
^(1,0)left(-1,m+frac32right)+frac14right)A^3 ,Gamma
left(m+frac32right)$$
$$prod_p=1^m a_2p+1=frac2 sqrt[12]2 Gamma (m+2) exp left(2 zeta
^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac14right)A^3$$
$$b_m=frac 12prod_p=1^m a_2pprod_p=1^m a_2p+1$$ $$b_m=frac2^frac 16sqrtpi Gamma (m+2) exp left(-2 zeta ^(1,0)(-1,m+1)+4
zeta ^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac12right)A^6 ,Gamma left(m+frac32right)$$
$$b_m=frac2^frac 16 sqrtpi , Gamma (m+2)A^4 ,H(m)^2,Gamma
left(m+frac32right)exp left(4 zeta ^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac13right)$$ where appears the hyperfactorial function.
Taking logarithms and using Stirling like approximations and then continuing with Taylor expansions using $b_m=e^log(b_m)$
$$b_m=frac2^frac 16 sqrt piA^6left(1+frac18 m-frac49384 m^2+frac1271024 m^3+Oleft(frac1m^4right) right)$$
$$colorbluelim_mto infty , b_m=frac2^frac 16 sqrt piA^6$$
answered Mar 14 at 8:43
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
Thank you for the answer @Claude Leibovic
$endgroup$
– coffeee
Mar 14 at 10:16
$begingroup$
@coffeee. You are very welcome ! Using Wolfram Alpha, type product of ((1 + 2*p)/(2 + 2*p))^(1 + 2*p)/(2^(2*p)*(p/(1 + 2*p))^(2*p)) from p=1 to infinity and you will get the result.
$endgroup$
– Claude Leibovici
Mar 14 at 10:22
$begingroup$
@coffeee. Take care : I could be wrong by a factor of two since I used $b_m=prod_p=1^m a_2pprod_p=1^m a_2p+1$ instead of $b_m=prod_p=1^m a_2pprod_p=colorred0^m a_2p+1$. I need to check again tomorrow morning.
$endgroup$
– Claude Leibovici
Mar 14 at 14:42
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3147649%2fclosed-form-of-prod-n-1-infty-left-fracnn1-right-1n-1n%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$displaystyle 1/prodlimits_n=1^2Nleft(fracnn+1right)^(-1)^n-1n = frac1sqrt2 left( frace^N/2N^-1/8 prodlimits_n=1^Nleft(1+frac12nright)^n
right)^4 left( frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n right)^-1 $
$displaystyle limlimits_Ntoinfty frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n = limlimits_Ntoinftyfrace^N N^-1/2 prodlimits_n=1^Nleft(1+frac1nright)^n = fracsqrt2pieenspaceenspace$ (e.g. by the Stirling formula)
The first calculation formula of Glaisher for the constant named after him (but written here more compact with products instead of series) is:
$$A=2^1/36pi^1/6left( limlimits_Ntoinfty frace^N/2N^-1/8 prodlimits_n=1^Nleft(1+frac12nright)^n
right)^2/3left( limlimits_Ntoinfty frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n right)^-1/3$$
(see Glaisher page 46 formula (7))
We potentiate this equation of Glaisher on both sides with 6 and multiply one time the left side with $fracsqrt2pie$ and the right with it’s product. After a few simple elementary conversions follows:
$displaystyle limlimits_Ntoinftyprodlimits_n=1^2Nleft(fracnn+1right)^(-1)^n-1n = 2^1/6pi^1/2eA^-6 approx 1.2157517513…$
$displaystyle limlimits_Ntoinftyprodlimits_n=1^2N+1left(fracnn+1right)^(-1)^n-1n = 2^1/6pi^1/2A^-6 approx 0.44725…$
answered Mar 14 at 16:08
user90369user90369
8,475925
$endgroup$
$begingroup$
thank you for showing the full answer
$endgroup$
– coffeee
Mar 15 at 20:31
$begingroup$
@coffeee : You are welcome, it was a pleasure. ;)
$endgroup$
– user90369
Mar 15 at 22:26
add a comment |
$begingroup$
$displaystyle 1/prodlimits_n=1^2Nleft(fracnn+1right)^(-1)^n-1n = frac1sqrt2 left( frace^N/2N^-1/8 prodlimits_n=1^Nleft(1+frac12nright)^n
right)^4 left( frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n right)^-1 $
$displaystyle limlimits_Ntoinfty frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n = limlimits_Ntoinftyfrace^N N^-1/2 prodlimits_n=1^Nleft(1+frac1nright)^n = fracsqrt2pieenspaceenspace$ (e.g. by the Stirling formula)
The first calculation formula of Glaisher for the constant named after him (but written here more compact with products instead of series) is:
$$A=2^1/36pi^1/6left( limlimits_Ntoinfty frace^N/2N^-1/8 prodlimits_n=1^Nleft(1+frac12nright)^n
right)^2/3left( limlimits_Ntoinfty frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n right)^-1/3$$
(see Glaisher page 46 formula (7))
We potentiate this equation of Glaisher on both sides with 6 and multiply one time the left side with $fracsqrt2pie$ and the right with it’s product. After a few simple elementary conversions follows:
$displaystyle limlimits_Ntoinftyprodlimits_n=1^2Nleft(fracnn+1right)^(-1)^n-1n = 2^1/6pi^1/2eA^-6 approx 1.2157517513…$
$displaystyle limlimits_Ntoinftyprodlimits_n=1^2N+1left(fracnn+1right)^(-1)^n-1n = 2^1/6pi^1/2A^-6 approx 0.44725…$
answered Mar 14 at 16:08
user90369user90369
8,475925
$endgroup$
$begingroup$
thank you for showing the full answer
$endgroup$
– coffeee
Mar 15 at 20:31
$begingroup$
@coffeee : You are welcome, it was a pleasure. ;)
$endgroup$
– user90369
Mar 15 at 22:26
add a comment |
$begingroup$
$displaystyle 1/prodlimits_n=1^2Nleft(fracnn+1right)^(-1)^n-1n = frac1sqrt2 left( frace^N/2N^-1/8 prodlimits_n=1^Nleft(1+frac12nright)^n
right)^4 left( frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n right)^-1 $
$displaystyle limlimits_Ntoinfty frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n = limlimits_Ntoinftyfrace^N N^-1/2 prodlimits_n=1^Nleft(1+frac1nright)^n = fracsqrt2pieenspaceenspace$ (e.g. by the Stirling formula)
The first calculation formula of Glaisher for the constant named after him (but written here more compact with products instead of series) is:
$$A=2^1/36pi^1/6left( limlimits_Ntoinfty frace^N/2N^-1/8 prodlimits_n=1^Nleft(1+frac12nright)^n
right)^2/3left( limlimits_Ntoinfty frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n right)^-1/3$$
(see Glaisher page 46 formula (7))
We potentiate this equation of Glaisher on both sides with 6 and multiply one time the left side with $fracsqrt2pie$ and the right with it’s product. After a few simple elementary conversions follows:
$displaystyle limlimits_Ntoinftyprodlimits_n=1^2Nleft(fracnn+1right)^(-1)^n-1n = 2^1/6pi^1/2eA^-6 approx 1.2157517513…$
$displaystyle limlimits_Ntoinftyprodlimits_n=1^2N+1left(fracnn+1right)^(-1)^n-1n = 2^1/6pi^1/2A^-6 approx 0.44725…$
answered Mar 14 at 16:08
user90369user90369
8,475925
$endgroup$
$displaystyle 1/prodlimits_n=1^2Nleft(fracnn+1right)^(-1)^n-1n = frac1sqrt2 left( frace^N/2N^-1/8 prodlimits_n=1^Nleft(1+frac12nright)^n
right)^4 left( frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n right)^-1 $
$displaystyle limlimits_Ntoinfty frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n = limlimits_Ntoinftyfrace^N N^-1/2 prodlimits_n=1^Nleft(1+frac1nright)^n = fracsqrt2pieenspaceenspace$ (e.g. by the Stirling formula)
The first calculation formula of Glaisher for the constant named after him (but written here more compact with products instead of series) is:
$$A=2^1/36pi^1/6left( limlimits_Ntoinfty frace^N/2N^-1/8 prodlimits_n=1^Nleft(1+frac12nright)^n
right)^2/3left( limlimits_Ntoinfty frace^2N(2N)^-1/2 prodlimits_n=1^2Nleft(1+frac1nright)^n right)^-1/3$$
(see Glaisher page 46 formula (7))
We potentiate this equation of Glaisher on both sides with 6 and multiply one time the left side with $fracsqrt2pie$ and the right with it’s product. After a few simple elementary conversions follows:
$displaystyle limlimits_Ntoinftyprodlimits_n=1^2Nleft(fracnn+1right)^(-1)^n-1n = 2^1/6pi^1/2eA^-6 approx 1.2157517513…$
$displaystyle limlimits_Ntoinftyprodlimits_n=1^2N+1left(fracnn+1right)^(-1)^n-1n = 2^1/6pi^1/2A^-6 approx 0.44725…$
answered Mar 14 at 16:08
user90369user90369
8,475925
edited Mar 14 at 18:42
answered Mar 14 at 16:08
user90369user90369
8,475925
answered Mar 14 at 16:08
user90369user90369
8,475925
answered Mar 14 at 16:08
user90369user90369
8,475925
8,475925
$begingroup$
thank you for showing the full answer
$endgroup$
– coffeee
Mar 15 at 20:31
$begingroup$
@coffeee : You are welcome, it was a pleasure. ;)
$endgroup$
– user90369
Mar 15 at 22:26
add a comment |
$begingroup$
thank you for showing the full answer
$endgroup$
– coffeee
Mar 15 at 20:31
$begingroup$
@coffeee : You are welcome, it was a pleasure. ;)
$endgroup$
– user90369
Mar 15 at 22:26
$begingroup$
thank you for showing the full answer
$endgroup$
– coffeee
Mar 15 at 20:31
$begingroup$
thank you for showing the full answer
$endgroup$
– coffeee
Mar 15 at 20:31
$begingroup$
@coffeee : You are welcome, it was a pleasure. ;)
$endgroup$
– user90369
Mar 15 at 22:26
$begingroup$
@coffeee : You are welcome, it was a pleasure. ;)
$endgroup$
– user90369
Mar 15 at 22:26
add a comment |
$begingroup$
Let $$a_n=left(fracnn+1right)^(-1)^n-1 n$$ then
$$a_2p= left(frac2p2 p+1right)^-2 pqquad textandqquad a_2p+1=left(frac2 p+12 p+2right)^2 p+1$$
Now, using a CAS,
$$prod_p=1^m a_2p=fracsqrt[12]2 sqrtpi exp left(-2 zeta ^(1,0)(-1,m+1)+2 zeta
^(1,0)left(-1,m+frac32right)+frac14right)A^3 ,Gamma
left(m+frac32right)$$
$$prod_p=1^m a_2p+1=frac2 sqrt[12]2 Gamma (m+2) exp left(2 zeta
^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac14right)A^3$$
$$b_m=frac 12prod_p=1^m a_2pprod_p=1^m a_2p+1$$ $$b_m=frac2^frac 16sqrtpi Gamma (m+2) exp left(-2 zeta ^(1,0)(-1,m+1)+4
zeta ^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac12right)A^6 ,Gamma left(m+frac32right)$$
$$b_m=frac2^frac 16 sqrtpi , Gamma (m+2)A^4 ,H(m)^2,Gamma
left(m+frac32right)exp left(4 zeta ^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac13right)$$ where appears the hyperfactorial function.
Taking logarithms and using Stirling like approximations and then continuing with Taylor expansions using $b_m=e^log(b_m)$
$$b_m=frac2^frac 16 sqrt piA^6left(1+frac18 m-frac49384 m^2+frac1271024 m^3+Oleft(frac1m^4right) right)$$
$$colorbluelim_mto infty , b_m=frac2^frac 16 sqrt piA^6$$
answered Mar 14 at 8:43
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
Thank you for the answer @Claude Leibovic
$endgroup$
– coffeee
Mar 14 at 10:16
$begingroup$
@coffeee. You are very welcome ! Using Wolfram Alpha, type product of ((1 + 2*p)/(2 + 2*p))^(1 + 2*p)/(2^(2*p)*(p/(1 + 2*p))^(2*p)) from p=1 to infinity and you will get the result.
$endgroup$
– Claude Leibovici
Mar 14 at 10:22
$begingroup$
@coffeee. Take care : I could be wrong by a factor of two since I used $b_m=prod_p=1^m a_2pprod_p=1^m a_2p+1$ instead of $b_m=prod_p=1^m a_2pprod_p=colorred0^m a_2p+1$. I need to check again tomorrow morning.
$endgroup$
– Claude Leibovici
Mar 14 at 14:42
add a comment |
$begingroup$
Let $$a_n=left(fracnn+1right)^(-1)^n-1 n$$ then
$$a_2p= left(frac2p2 p+1right)^-2 pqquad textandqquad a_2p+1=left(frac2 p+12 p+2right)^2 p+1$$
Now, using a CAS,
$$prod_p=1^m a_2p=fracsqrt[12]2 sqrtpi exp left(-2 zeta ^(1,0)(-1,m+1)+2 zeta
^(1,0)left(-1,m+frac32right)+frac14right)A^3 ,Gamma
left(m+frac32right)$$
$$prod_p=1^m a_2p+1=frac2 sqrt[12]2 Gamma (m+2) exp left(2 zeta
^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac14right)A^3$$
$$b_m=frac 12prod_p=1^m a_2pprod_p=1^m a_2p+1$$ $$b_m=frac2^frac 16sqrtpi Gamma (m+2) exp left(-2 zeta ^(1,0)(-1,m+1)+4
zeta ^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac12right)A^6 ,Gamma left(m+frac32right)$$
$$b_m=frac2^frac 16 sqrtpi , Gamma (m+2)A^4 ,H(m)^2,Gamma
left(m+frac32right)exp left(4 zeta ^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac13right)$$ where appears the hyperfactorial function.
Taking logarithms and using Stirling like approximations and then continuing with Taylor expansions using $b_m=e^log(b_m)$
$$b_m=frac2^frac 16 sqrt piA^6left(1+frac18 m-frac49384 m^2+frac1271024 m^3+Oleft(frac1m^4right) right)$$
$$colorbluelim_mto infty , b_m=frac2^frac 16 sqrt piA^6$$
answered Mar 14 at 8:43
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
Thank you for the answer @Claude Leibovic
$endgroup$
– coffeee
Mar 14 at 10:16
$begingroup$
@coffeee. You are very welcome ! Using Wolfram Alpha, type product of ((1 + 2*p)/(2 + 2*p))^(1 + 2*p)/(2^(2*p)*(p/(1 + 2*p))^(2*p)) from p=1 to infinity and you will get the result.
$endgroup$
– Claude Leibovici
Mar 14 at 10:22
$begingroup$
@coffeee. Take care : I could be wrong by a factor of two since I used $b_m=prod_p=1^m a_2pprod_p=1^m a_2p+1$ instead of $b_m=prod_p=1^m a_2pprod_p=colorred0^m a_2p+1$. I need to check again tomorrow morning.
$endgroup$
– Claude Leibovici
Mar 14 at 14:42
add a comment |
$begingroup$
Let $$a_n=left(fracnn+1right)^(-1)^n-1 n$$ then
$$a_2p= left(frac2p2 p+1right)^-2 pqquad textandqquad a_2p+1=left(frac2 p+12 p+2right)^2 p+1$$
Now, using a CAS,
$$prod_p=1^m a_2p=fracsqrt[12]2 sqrtpi exp left(-2 zeta ^(1,0)(-1,m+1)+2 zeta
^(1,0)left(-1,m+frac32right)+frac14right)A^3 ,Gamma
left(m+frac32right)$$
$$prod_p=1^m a_2p+1=frac2 sqrt[12]2 Gamma (m+2) exp left(2 zeta
^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac14right)A^3$$
$$b_m=frac 12prod_p=1^m a_2pprod_p=1^m a_2p+1$$ $$b_m=frac2^frac 16sqrtpi Gamma (m+2) exp left(-2 zeta ^(1,0)(-1,m+1)+4
zeta ^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac12right)A^6 ,Gamma left(m+frac32right)$$
$$b_m=frac2^frac 16 sqrtpi , Gamma (m+2)A^4 ,H(m)^2,Gamma
left(m+frac32right)exp left(4 zeta ^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac13right)$$ where appears the hyperfactorial function.
Taking logarithms and using Stirling like approximations and then continuing with Taylor expansions using $b_m=e^log(b_m)$
$$b_m=frac2^frac 16 sqrt piA^6left(1+frac18 m-frac49384 m^2+frac1271024 m^3+Oleft(frac1m^4right) right)$$
$$colorbluelim_mto infty , b_m=frac2^frac 16 sqrt piA^6$$
answered Mar 14 at 8:43
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
Let $$a_n=left(fracnn+1right)^(-1)^n-1 n$$ then
$$a_2p= left(frac2p2 p+1right)^-2 pqquad textandqquad a_2p+1=left(frac2 p+12 p+2right)^2 p+1$$
Now, using a CAS,
$$prod_p=1^m a_2p=fracsqrt[12]2 sqrtpi exp left(-2 zeta ^(1,0)(-1,m+1)+2 zeta
^(1,0)left(-1,m+frac32right)+frac14right)A^3 ,Gamma
left(m+frac32right)$$
$$prod_p=1^m a_2p+1=frac2 sqrt[12]2 Gamma (m+2) exp left(2 zeta
^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac14right)A^3$$
$$b_m=frac 12prod_p=1^m a_2pprod_p=1^m a_2p+1$$ $$b_m=frac2^frac 16sqrtpi Gamma (m+2) exp left(-2 zeta ^(1,0)(-1,m+1)+4
zeta ^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac12right)A^6 ,Gamma left(m+frac32right)$$
$$b_m=frac2^frac 16 sqrtpi , Gamma (m+2)A^4 ,H(m)^2,Gamma
left(m+frac32right)exp left(4 zeta ^(1,0)left(-1,m+frac32right)-2 zeta
^(1,0)(-1,m+2)+frac13right)$$ where appears the hyperfactorial function.
Taking logarithms and using Stirling like approximations and then continuing with Taylor expansions using $b_m=e^log(b_m)$
$$b_m=frac2^frac 16 sqrt piA^6left(1+frac18 m-frac49384 m^2+frac1271024 m^3+Oleft(frac1m^4right) right)$$
$$colorbluelim_mto infty , b_m=frac2^frac 16 sqrt piA^6$$
answered Mar 14 at 8:43
Claude LeiboviciClaude Leibovici
124k1158135
edited Mar 15 at 3:44
answered Mar 14 at 8:43
Claude LeiboviciClaude Leibovici
124k1158135
answered Mar 14 at 8:43
Claude LeiboviciClaude Leibovici
124k1158135
answered Mar 14 at 8:43
Claude LeiboviciClaude Leibovici
124k1158135
124k1158135
$begingroup$
Thank you for the answer @Claude Leibovic
$endgroup$
– coffeee
Mar 14 at 10:16
$begingroup$
@coffeee. You are very welcome ! Using Wolfram Alpha, type product of ((1 + 2*p)/(2 + 2*p))^(1 + 2*p)/(2^(2*p)*(p/(1 + 2*p))^(2*p)) from p=1 to infinity and you will get the result.
$endgroup$
– Claude Leibovici
Mar 14 at 10:22
$begingroup$
@coffeee. Take care : I could be wrong by a factor of two since I used $b_m=prod_p=1^m a_2pprod_p=1^m a_2p+1$ instead of $b_m=prod_p=1^m a_2pprod_p=colorred0^m a_2p+1$. I need to check again tomorrow morning.
$endgroup$
– Claude Leibovici
Mar 14 at 14:42
add a comment |
$begingroup$
Thank you for the answer @Claude Leibovic
$endgroup$
– coffeee
Mar 14 at 10:16
$begingroup$
@coffeee. You are very welcome ! Using Wolfram Alpha, type product of ((1 + 2*p)/(2 + 2*p))^(1 + 2*p)/(2^(2*p)*(p/(1 + 2*p))^(2*p)) from p=1 to infinity and you will get the result.
$endgroup$
– Claude Leibovici
Mar 14 at 10:22
$begingroup$
@coffeee. Take care : I could be wrong by a factor of two since I used $b_m=prod_p=1^m a_2pprod_p=1^m a_2p+1$ instead of $b_m=prod_p=1^m a_2pprod_p=colorred0^m a_2p+1$. I need to check again tomorrow morning.
$endgroup$
– Claude Leibovici
Mar 14 at 14:42
$begingroup$
Thank you for the answer @Claude Leibovic
$endgroup$
– coffeee
Mar 14 at 10:16
$begingroup$
Thank you for the answer @Claude Leibovic
$endgroup$
– coffeee
Mar 14 at 10:16
$begingroup$
@coffeee. You are very welcome ! Using Wolfram Alpha, type product of ((1 + 2*p)/(2 + 2*p))^(1 + 2*p)/(2^(2*p)*(p/(1 + 2*p))^(2*p)) from p=1 to infinity and you will get the result.
$endgroup$
– Claude Leibovici
Mar 14 at 10:22
$begingroup$
@coffeee. You are very welcome ! Using Wolfram Alpha, type product of ((1 + 2*p)/(2 + 2*p))^(1 + 2*p)/(2^(2*p)*(p/(1 + 2*p))^(2*p)) from p=1 to infinity and you will get the result.
$endgroup$
– Claude Leibovici
Mar 14 at 10:22
$begingroup$
@coffeee. Take care : I could be wrong by a factor of two since I used $b_m=prod_p=1^m a_2pprod_p=1^m a_2p+1$ instead of $b_m=prod_p=1^m a_2pprod_p=colorred0^m a_2p+1$. I need to check again tomorrow morning.
$endgroup$
– Claude Leibovici
Mar 14 at 14:42
$begingroup$
@coffeee. Take care : I could be wrong by a factor of two since I used $b_m=prod_p=1^m a_2pprod_p=1^m a_2p+1$ instead of $b_m=prod_p=1^m a_2pprod_p=colorred0^m a_2p+1$. I need to check again tomorrow morning.
$endgroup$
– Claude Leibovici
Mar 14 at 14:42
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3147649%2fclosed-form-of-prod-n-1-infty-left-fracnn1-right-1n-1n%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Is it convergent? I got the product to be $$frac 3^5 cdot 5^9 cdot 7^13 cdots 2^3 cdot 4^7 cdot 6^11 cdots$$ which is definitely divergent.
$endgroup$
– Dbchatto67
Mar 14 at 7:32
$begingroup$
There are two limits possible, please see my answer.
$endgroup$
– user90369
Mar 14 at 16:51