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Is WolframAlpha wrong about $ lim_xtoinfty(x!)^1/x-fracxe $?
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I need some help evaluating this limit... Wolfram says it blows up to infinity but I don't think so. I just can't prove it yet.
$$ lim_xtoinfty(x!)^1/x-fracxe $$
calculus limits
edited Aug 1 at 17:33
amWhy
189k25219431
asked Aug 1 at 17:25
Blake Steines
574
 |Â
show 3 more comments
up vote
1
down vote
favorite
I need some help evaluating this limit... Wolfram says it blows up to infinity but I don't think so. I just can't prove it yet.
$$ lim_xtoinfty(x!)^1/x-fracxe $$
calculus limits
edited Aug 1 at 17:33
amWhy
189k25219431
asked Aug 1 at 17:25
Blake Steines
574
1
Have you tried Stirling's formula?
– Lord Shark the Unknown
Aug 1 at 17:32
If the limit is actually $infty$, the sequence diverges very slow. With PARI/GP, I got about value $3$ for $x=2000000$
– Peter
Aug 1 at 17:35
$log(n!)approx nlog n -n$
– saulspatz
Aug 1 at 17:41
1
If Maple and I did not make any mistake you have $$Gamma(x+1)^1/x-x/e = frac12eln(2pi x) + O(1/x), quad xrightarrow infty$$
– gammatester
Aug 1 at 17:44
@saulspatz unless I'm mistaken, your formula would imply that the limit does not blow up to infinity.
– Blake Steines
Aug 1 at 17:45
 |Â
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need some help evaluating this limit... Wolfram says it blows up to infinity but I don't think so. I just can't prove it yet.
$$ lim_xtoinfty(x!)^1/x-fracxe $$
calculus limits
edited Aug 1 at 17:33
amWhy
189k25219431
asked Aug 1 at 17:25
Blake Steines
574
I need some help evaluating this limit... Wolfram says it blows up to infinity but I don't think so. I just can't prove it yet.
$$ lim_xtoinfty(x!)^1/x-fracxe $$
calculus limits
edited Aug 1 at 17:33
amWhy
189k25219431
asked Aug 1 at 17:25
Blake Steines
574
edited Aug 1 at 17:33
amWhy
189k25219431
edited Aug 1 at 17:33
amWhy
189k25219431
edited Aug 1 at 17:33
amWhy
189k25219431
189k25219431
asked Aug 1 at 17:25
Blake Steines
574
asked Aug 1 at 17:25
Blake Steines
574
asked Aug 1 at 17:25
Blake Steines
574
574
1
Have you tried Stirling's formula?
– Lord Shark the Unknown
Aug 1 at 17:32
If the limit is actually $infty$, the sequence diverges very slow. With PARI/GP, I got about value $3$ for $x=2000000$
– Peter
Aug 1 at 17:35
$log(n!)approx nlog n -n$
– saulspatz
Aug 1 at 17:41
1
If Maple and I did not make any mistake you have $$Gamma(x+1)^1/x-x/e = frac12eln(2pi x) + O(1/x), quad xrightarrow infty$$
– gammatester
Aug 1 at 17:44
@saulspatz unless I'm mistaken, your formula would imply that the limit does not blow up to infinity.
– Blake Steines
Aug 1 at 17:45
 |Â
show 3 more comments
1
Have you tried Stirling's formula?
– Lord Shark the Unknown
Aug 1 at 17:32
If the limit is actually $infty$, the sequence diverges very slow. With PARI/GP, I got about value $3$ for $x=2000000$
– Peter
Aug 1 at 17:35
$log(n!)approx nlog n -n$
– saulspatz
Aug 1 at 17:41
1
If Maple and I did not make any mistake you have $$Gamma(x+1)^1/x-x/e = frac12eln(2pi x) + O(1/x), quad xrightarrow infty$$
– gammatester
Aug 1 at 17:44
@saulspatz unless I'm mistaken, your formula would imply that the limit does not blow up to infinity.
– Blake Steines
Aug 1 at 17:45
1
1
Have you tried Stirling's formula?
– Lord Shark the Unknown
Aug 1 at 17:32
Have you tried Stirling's formula?
– Lord Shark the Unknown
Aug 1 at 17:32
If the limit is actually $infty$, the sequence diverges very slow. With PARI/GP, I got about value $3$ for $x=2000000$
– Peter
Aug 1 at 17:35
If the limit is actually $infty$, the sequence diverges very slow. With PARI/GP, I got about value $3$ for $x=2000000$
– Peter
Aug 1 at 17:35
$log(n!)approx nlog n -n$
– saulspatz
Aug 1 at 17:41
$log(n!)approx nlog n -n$
– saulspatz
Aug 1 at 17:41
1
1
If Maple and I did not make any mistake you have $$Gamma(x+1)^1/x-x/e = frac12eln(2pi x) + O(1/x), quad xrightarrow infty$$
– gammatester
Aug 1 at 17:44
If Maple and I did not make any mistake you have $$Gamma(x+1)^1/x-x/e = frac12eln(2pi x) + O(1/x), quad xrightarrow infty$$
– gammatester
Aug 1 at 17:44
@saulspatz unless I'm mistaken, your formula would imply that the limit does not blow up to infinity.
– Blake Steines
Aug 1 at 17:45
@saulspatz unless I'm mistaken, your formula would imply that the limit does not blow up to infinity.
– Blake Steines
Aug 1 at 17:45
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
5
down vote
It diverges, albeit slowly.
By Stirling's Approximation, the function approaches
$$f(x)=(2pi x)^1/xleft( frac xeright)-frac xe=left(frac xeright)left( (2pi x)^1/x-1right).$$
Now it is true that $(2pi x)^1/x$ goes to $1$ as $xtoinfty$, but we need more information. Using L'Hospital's rule,
beginalign
lim_xtoinftyfrac(2pi x)^1/x-1e/x&=lim_xtoinftyfrac-(2 pi x)^1/x x^-2 (log (2 pi x)-1)-ex^-2\
&=lim_xtoinftye^-1(2 pi x)^1/x (log (2 pi x)-1)\
&=infty
endalign
since the factor $e^-1(2 pi x)^1/x$ is bounded and nonzero and $log(2pi x)-1toinfty$.
As an aside, check out the harmonic prime series $1/2+1/3+1/5+1/7+cdots$. This diverges, but so slowly that the first million primes don't even make it past $3$.
edited Aug 2 at 13:10
Moo
4,7233920
answered Aug 1 at 17:47
Elliot G
9,66521645
What then, is incorrect about my logic here (using n! ~ nlogn - n): (x!)^(1/x) = x/e, then (1/x)*(xlogx -x) = ln(x) - 1, then ln(x) -1 = ln(x) -1, so the limits should be equal?
– Blake Steines
Aug 1 at 17:53
2
That the logarithms approach the same number tells you that their quotient approaches $1,$ not that their difference approaches $0$.
– saulspatz
Aug 1 at 18:00
** using log(n!) = nlogn - n
– Blake Steines
Aug 1 at 18:00
First, I think you're missing an extra $log$ on the left. So if you have $(x!)^1/x=x/e$, then taking $log$ of both sides would give $frac 1x log(xlog x-x)=log x-1$, or $log(xlog x-x)=xlog x-1$. But also it makes no sense to begin with the assumption $(x!)^1/x=x/e$. The solution here would only show you where the function has a root.
– Elliot G
Aug 1 at 18:01
2
I upvoted this, but now I have doubts. Stirling's approximation is asymptotic, so is it valid to use in in computing a difference?
– saulspatz
Aug 1 at 18:08
 |Â
show 1 more comment
up vote
2
down vote
I used a simple one-liner and asked Maple 7 to simplify(series(GAMMA(x+1)^(1/x)-x/exp(1), x=infinity,3)); and the answer is 1/2*(ln(2)+ln(Pi)+ln(x)+2*O(1/x)*exp(1))*exp(-1), this can be converted to mathematical notation as:
$$Gammaleft(x+1right)^1/x-fracxe = frac12eln(2pi x) + Oleft(frac1xright), quad xrightarrow infty$$
answered Aug 1 at 19:52
gammatester
15.7k21429
Thanks. I wish we could see the underlying reasoning.
– saulspatz
Aug 1 at 20:01
I suppose that any expansion of $log(p!)$ followed by Taylor series would confirm it. By the way, $+1$. Cheers.
– Claude Leibovici
Aug 2 at 10:50
add a comment |Â
up vote
2
down vote
In the same spirit as in previous answers, considering $$A=(x!)^frac1x-fracxe=B-fracxe$$
$$B=(x!)^frac1ximplies log(B)=frac1xlog(x!)$$ Now, using Stirling approximation
$$log(B)=frac1xleft(x (log (x)-1)+frac12 left(log (2 pi )+log
left(xright)right)+frac112
x+Oleft(frac1x^3right)right)$$
$$log(B)= (log (x)-1)+frac12x left(log (2 pi )+log
left(xright)right)+Oleft(frac1x^2right)$$ Continuing with Taylor
$$B=e^log(B)=fracxe+fraclog (2 pi )+log left(xright)2 e+Oleft(frac1xright)$$
$$A=fraclog (2 pi x)2 e+Oleft(frac1xright)tag 1$$ Pushing further the expansion of $B$, you would get
$$A=fraclog (2 pi x)2 e+frac3 log ^2left(2 pi xright)+224 e x+Oleft(frac1x^2right)tag 2$$ For illustration purposes, let $x=10^k$ and computing
$$left(
beginarraycccc
k & (1) & (2) & textexact \
1 & 0.761595453 & 0.843495044 & 0.849934276 \
2 & 1.185132311 & 1.204528536 & 1.204745228 \
3 & 1.608669170 & 1.612217034 & 1.612222301 \
4 & 2.032206029 & 2.032770401 & 2.032770506
endarray
right)$$
answered Aug 2 at 4:37
Claude Leibovici
111k1054126
(+1) So you have confirmed that the Maple expression is correct.
– gammatester
Aug 2 at 10:27
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
It diverges, albeit slowly.
By Stirling's Approximation, the function approaches
$$f(x)=(2pi x)^1/xleft( frac xeright)-frac xe=left(frac xeright)left( (2pi x)^1/x-1right).$$
Now it is true that $(2pi x)^1/x$ goes to $1$ as $xtoinfty$, but we need more information. Using L'Hospital's rule,
beginalign
lim_xtoinftyfrac(2pi x)^1/x-1e/x&=lim_xtoinftyfrac-(2 pi x)^1/x x^-2 (log (2 pi x)-1)-ex^-2\
&=lim_xtoinftye^-1(2 pi x)^1/x (log (2 pi x)-1)\
&=infty
endalign
since the factor $e^-1(2 pi x)^1/x$ is bounded and nonzero and $log(2pi x)-1toinfty$.
As an aside, check out the harmonic prime series $1/2+1/3+1/5+1/7+cdots$. This diverges, but so slowly that the first million primes don't even make it past $3$.
edited Aug 2 at 13:10
Moo
4,7233920
answered Aug 1 at 17:47
Elliot G
9,66521645
What then, is incorrect about my logic here (using n! ~ nlogn - n): (x!)^(1/x) = x/e, then (1/x)*(xlogx -x) = ln(x) - 1, then ln(x) -1 = ln(x) -1, so the limits should be equal?
– Blake Steines
Aug 1 at 17:53
2
That the logarithms approach the same number tells you that their quotient approaches $1,$ not that their difference approaches $0$.
– saulspatz
Aug 1 at 18:00
** using log(n!) = nlogn - n
– Blake Steines
Aug 1 at 18:00
First, I think you're missing an extra $log$ on the left. So if you have $(x!)^1/x=x/e$, then taking $log$ of both sides would give $frac 1x log(xlog x-x)=log x-1$, or $log(xlog x-x)=xlog x-1$. But also it makes no sense to begin with the assumption $(x!)^1/x=x/e$. The solution here would only show you where the function has a root.
– Elliot G
Aug 1 at 18:01
2
I upvoted this, but now I have doubts. Stirling's approximation is asymptotic, so is it valid to use in in computing a difference?
– saulspatz
Aug 1 at 18:08
 |Â
show 1 more comment
up vote
5
down vote
It diverges, albeit slowly.
By Stirling's Approximation, the function approaches
$$f(x)=(2pi x)^1/xleft( frac xeright)-frac xe=left(frac xeright)left( (2pi x)^1/x-1right).$$
Now it is true that $(2pi x)^1/x$ goes to $1$ as $xtoinfty$, but we need more information. Using L'Hospital's rule,
beginalign
lim_xtoinftyfrac(2pi x)^1/x-1e/x&=lim_xtoinftyfrac-(2 pi x)^1/x x^-2 (log (2 pi x)-1)-ex^-2\
&=lim_xtoinftye^-1(2 pi x)^1/x (log (2 pi x)-1)\
&=infty
endalign
since the factor $e^-1(2 pi x)^1/x$ is bounded and nonzero and $log(2pi x)-1toinfty$.
As an aside, check out the harmonic prime series $1/2+1/3+1/5+1/7+cdots$. This diverges, but so slowly that the first million primes don't even make it past $3$.
edited Aug 2 at 13:10
Moo
4,7233920
answered Aug 1 at 17:47
Elliot G
9,66521645
What then, is incorrect about my logic here (using n! ~ nlogn - n): (x!)^(1/x) = x/e, then (1/x)*(xlogx -x) = ln(x) - 1, then ln(x) -1 = ln(x) -1, so the limits should be equal?
– Blake Steines
Aug 1 at 17:53
2
That the logarithms approach the same number tells you that their quotient approaches $1,$ not that their difference approaches $0$.
– saulspatz
Aug 1 at 18:00
** using log(n!) = nlogn - n
– Blake Steines
Aug 1 at 18:00
First, I think you're missing an extra $log$ on the left. So if you have $(x!)^1/x=x/e$, then taking $log$ of both sides would give $frac 1x log(xlog x-x)=log x-1$, or $log(xlog x-x)=xlog x-1$. But also it makes no sense to begin with the assumption $(x!)^1/x=x/e$. The solution here would only show you where the function has a root.
– Elliot G
Aug 1 at 18:01
2
I upvoted this, but now I have doubts. Stirling's approximation is asymptotic, so is it valid to use in in computing a difference?
– saulspatz
Aug 1 at 18:08
 |Â
show 1 more comment
up vote
5
down vote
up vote
5
down vote
It diverges, albeit slowly.
By Stirling's Approximation, the function approaches
$$f(x)=(2pi x)^1/xleft( frac xeright)-frac xe=left(frac xeright)left( (2pi x)^1/x-1right).$$
Now it is true that $(2pi x)^1/x$ goes to $1$ as $xtoinfty$, but we need more information. Using L'Hospital's rule,
beginalign
lim_xtoinftyfrac(2pi x)^1/x-1e/x&=lim_xtoinftyfrac-(2 pi x)^1/x x^-2 (log (2 pi x)-1)-ex^-2\
&=lim_xtoinftye^-1(2 pi x)^1/x (log (2 pi x)-1)\
&=infty
endalign
since the factor $e^-1(2 pi x)^1/x$ is bounded and nonzero and $log(2pi x)-1toinfty$.
As an aside, check out the harmonic prime series $1/2+1/3+1/5+1/7+cdots$. This diverges, but so slowly that the first million primes don't even make it past $3$.
edited Aug 2 at 13:10
Moo
4,7233920
answered Aug 1 at 17:47
Elliot G
9,66521645
It diverges, albeit slowly.
By Stirling's Approximation, the function approaches
$$f(x)=(2pi x)^1/xleft( frac xeright)-frac xe=left(frac xeright)left( (2pi x)^1/x-1right).$$
Now it is true that $(2pi x)^1/x$ goes to $1$ as $xtoinfty$, but we need more information. Using L'Hospital's rule,
beginalign
lim_xtoinftyfrac(2pi x)^1/x-1e/x&=lim_xtoinftyfrac-(2 pi x)^1/x x^-2 (log (2 pi x)-1)-ex^-2\
&=lim_xtoinftye^-1(2 pi x)^1/x (log (2 pi x)-1)\
&=infty
endalign
since the factor $e^-1(2 pi x)^1/x$ is bounded and nonzero and $log(2pi x)-1toinfty$.
As an aside, check out the harmonic prime series $1/2+1/3+1/5+1/7+cdots$. This diverges, but so slowly that the first million primes don't even make it past $3$.
edited Aug 2 at 13:10
Moo
4,7233920
answered Aug 1 at 17:47
Elliot G
9,66521645
edited Aug 2 at 13:10
Moo
4,7233920
edited Aug 2 at 13:10
Moo
4,7233920
edited Aug 2 at 13:10
Moo
4,7233920
4,7233920
answered Aug 1 at 17:47
Elliot G
9,66521645
answered Aug 1 at 17:47
Elliot G
9,66521645
answered Aug 1 at 17:47
Elliot G
9,66521645
9,66521645
What then, is incorrect about my logic here (using n! ~ nlogn - n): (x!)^(1/x) = x/e, then (1/x)*(xlogx -x) = ln(x) - 1, then ln(x) -1 = ln(x) -1, so the limits should be equal?
– Blake Steines
Aug 1 at 17:53
2
That the logarithms approach the same number tells you that their quotient approaches $1,$ not that their difference approaches $0$.
– saulspatz
Aug 1 at 18:00
** using log(n!) = nlogn - n
– Blake Steines
Aug 1 at 18:00
First, I think you're missing an extra $log$ on the left. So if you have $(x!)^1/x=x/e$, then taking $log$ of both sides would give $frac 1x log(xlog x-x)=log x-1$, or $log(xlog x-x)=xlog x-1$. But also it makes no sense to begin with the assumption $(x!)^1/x=x/e$. The solution here would only show you where the function has a root.
– Elliot G
Aug 1 at 18:01
2
I upvoted this, but now I have doubts. Stirling's approximation is asymptotic, so is it valid to use in in computing a difference?
– saulspatz
Aug 1 at 18:08
 |Â
show 1 more comment
What then, is incorrect about my logic here (using n! ~ nlogn - n): (x!)^(1/x) = x/e, then (1/x)*(xlogx -x) = ln(x) - 1, then ln(x) -1 = ln(x) -1, so the limits should be equal?
– Blake Steines
Aug 1 at 17:53
2
That the logarithms approach the same number tells you that their quotient approaches $1,$ not that their difference approaches $0$.
– saulspatz
Aug 1 at 18:00
** using log(n!) = nlogn - n
– Blake Steines
Aug 1 at 18:00
First, I think you're missing an extra $log$ on the left. So if you have $(x!)^1/x=x/e$, then taking $log$ of both sides would give $frac 1x log(xlog x-x)=log x-1$, or $log(xlog x-x)=xlog x-1$. But also it makes no sense to begin with the assumption $(x!)^1/x=x/e$. The solution here would only show you where the function has a root.
– Elliot G
Aug 1 at 18:01
2
I upvoted this, but now I have doubts. Stirling's approximation is asymptotic, so is it valid to use in in computing a difference?
– saulspatz
Aug 1 at 18:08
What then, is incorrect about my logic here (using n! ~ nlogn - n): (x!)^(1/x) = x/e, then (1/x)*(xlogx -x) = ln(x) - 1, then ln(x) -1 = ln(x) -1, so the limits should be equal?
– Blake Steines
Aug 1 at 17:53
What then, is incorrect about my logic here (using n! ~ nlogn - n): (x!)^(1/x) = x/e, then (1/x)*(xlogx -x) = ln(x) - 1, then ln(x) -1 = ln(x) -1, so the limits should be equal?
– Blake Steines
Aug 1 at 17:53
2
2
That the logarithms approach the same number tells you that their quotient approaches $1,$ not that their difference approaches $0$.
– saulspatz
Aug 1 at 18:00
That the logarithms approach the same number tells you that their quotient approaches $1,$ not that their difference approaches $0$.
– saulspatz
Aug 1 at 18:00
** using log(n!) = nlogn - n
– Blake Steines
Aug 1 at 18:00
** using log(n!) = nlogn - n
– Blake Steines
Aug 1 at 18:00
First, I think you're missing an extra $log$ on the left. So if you have $(x!)^1/x=x/e$, then taking $log$ of both sides would give $frac 1x log(xlog x-x)=log x-1$, or $log(xlog x-x)=xlog x-1$. But also it makes no sense to begin with the assumption $(x!)^1/x=x/e$. The solution here would only show you where the function has a root.
– Elliot G
Aug 1 at 18:01
First, I think you're missing an extra $log$ on the left. So if you have $(x!)^1/x=x/e$, then taking $log$ of both sides would give $frac 1x log(xlog x-x)=log x-1$, or $log(xlog x-x)=xlog x-1$. But also it makes no sense to begin with the assumption $(x!)^1/x=x/e$. The solution here would only show you where the function has a root.
– Elliot G
Aug 1 at 18:01
2
2
I upvoted this, but now I have doubts. Stirling's approximation is asymptotic, so is it valid to use in in computing a difference?
– saulspatz
Aug 1 at 18:08
I upvoted this, but now I have doubts. Stirling's approximation is asymptotic, so is it valid to use in in computing a difference?
– saulspatz
Aug 1 at 18:08
 |Â
show 1 more comment
up vote
2
down vote
I used a simple one-liner and asked Maple 7 to simplify(series(GAMMA(x+1)^(1/x)-x/exp(1), x=infinity,3)); and the answer is 1/2*(ln(2)+ln(Pi)+ln(x)+2*O(1/x)*exp(1))*exp(-1), this can be converted to mathematical notation as:
$$Gammaleft(x+1right)^1/x-fracxe = frac12eln(2pi x) + Oleft(frac1xright), quad xrightarrow infty$$
answered Aug 1 at 19:52
gammatester
15.7k21429
Thanks. I wish we could see the underlying reasoning.
– saulspatz
Aug 1 at 20:01
I suppose that any expansion of $log(p!)$ followed by Taylor series would confirm it. By the way, $+1$. Cheers.
– Claude Leibovici
Aug 2 at 10:50
add a comment |Â
up vote
2
down vote
I used a simple one-liner and asked Maple 7 to simplify(series(GAMMA(x+1)^(1/x)-x/exp(1), x=infinity,3)); and the answer is 1/2*(ln(2)+ln(Pi)+ln(x)+2*O(1/x)*exp(1))*exp(-1), this can be converted to mathematical notation as:
$$Gammaleft(x+1right)^1/x-fracxe = frac12eln(2pi x) + Oleft(frac1xright), quad xrightarrow infty$$
answered Aug 1 at 19:52
gammatester
15.7k21429
Thanks. I wish we could see the underlying reasoning.
– saulspatz
Aug 1 at 20:01
I suppose that any expansion of $log(p!)$ followed by Taylor series would confirm it. By the way, $+1$. Cheers.
– Claude Leibovici
Aug 2 at 10:50
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I used a simple one-liner and asked Maple 7 to simplify(series(GAMMA(x+1)^(1/x)-x/exp(1), x=infinity,3)); and the answer is 1/2*(ln(2)+ln(Pi)+ln(x)+2*O(1/x)*exp(1))*exp(-1), this can be converted to mathematical notation as:
$$Gammaleft(x+1right)^1/x-fracxe = frac12eln(2pi x) + Oleft(frac1xright), quad xrightarrow infty$$
answered Aug 1 at 19:52
gammatester
15.7k21429
I used a simple one-liner and asked Maple 7 to simplify(series(GAMMA(x+1)^(1/x)-x/exp(1), x=infinity,3)); and the answer is 1/2*(ln(2)+ln(Pi)+ln(x)+2*O(1/x)*exp(1))*exp(-1), this can be converted to mathematical notation as:
$$Gammaleft(x+1right)^1/x-fracxe = frac12eln(2pi x) + Oleft(frac1xright), quad xrightarrow infty$$
answered Aug 1 at 19:52
gammatester
15.7k21429
answered Aug 1 at 19:52
gammatester
15.7k21429
answered Aug 1 at 19:52
gammatester
15.7k21429
answered Aug 1 at 19:52
gammatester
15.7k21429
15.7k21429
Thanks. I wish we could see the underlying reasoning.
– saulspatz
Aug 1 at 20:01
I suppose that any expansion of $log(p!)$ followed by Taylor series would confirm it. By the way, $+1$. Cheers.
– Claude Leibovici
Aug 2 at 10:50
add a comment |Â
Thanks. I wish we could see the underlying reasoning.
– saulspatz
Aug 1 at 20:01
I suppose that any expansion of $log(p!)$ followed by Taylor series would confirm it. By the way, $+1$. Cheers.
– Claude Leibovici
Aug 2 at 10:50
Thanks. I wish we could see the underlying reasoning.
– saulspatz
Aug 1 at 20:01
Thanks. I wish we could see the underlying reasoning.
– saulspatz
Aug 1 at 20:01
I suppose that any expansion of $log(p!)$ followed by Taylor series would confirm it. By the way, $+1$. Cheers.
– Claude Leibovici
Aug 2 at 10:50
I suppose that any expansion of $log(p!)$ followed by Taylor series would confirm it. By the way, $+1$. Cheers.
– Claude Leibovici
Aug 2 at 10:50
add a comment |Â
up vote
2
down vote
In the same spirit as in previous answers, considering $$A=(x!)^frac1x-fracxe=B-fracxe$$
$$B=(x!)^frac1ximplies log(B)=frac1xlog(x!)$$ Now, using Stirling approximation
$$log(B)=frac1xleft(x (log (x)-1)+frac12 left(log (2 pi )+log
left(xright)right)+frac112
x+Oleft(frac1x^3right)right)$$
$$log(B)= (log (x)-1)+frac12x left(log (2 pi )+log
left(xright)right)+Oleft(frac1x^2right)$$ Continuing with Taylor
$$B=e^log(B)=fracxe+fraclog (2 pi )+log left(xright)2 e+Oleft(frac1xright)$$
$$A=fraclog (2 pi x)2 e+Oleft(frac1xright)tag 1$$ Pushing further the expansion of $B$, you would get
$$A=fraclog (2 pi x)2 e+frac3 log ^2left(2 pi xright)+224 e x+Oleft(frac1x^2right)tag 2$$ For illustration purposes, let $x=10^k$ and computing
$$left(
beginarraycccc
k & (1) & (2) & textexact \
1 & 0.761595453 & 0.843495044 & 0.849934276 \
2 & 1.185132311 & 1.204528536 & 1.204745228 \
3 & 1.608669170 & 1.612217034 & 1.612222301 \
4 & 2.032206029 & 2.032770401 & 2.032770506
endarray
right)$$
answered Aug 2 at 4:37
Claude Leibovici
111k1054126
(+1) So you have confirmed that the Maple expression is correct.
– gammatester
Aug 2 at 10:27
add a comment |Â
up vote
2
down vote
In the same spirit as in previous answers, considering $$A=(x!)^frac1x-fracxe=B-fracxe$$
$$B=(x!)^frac1ximplies log(B)=frac1xlog(x!)$$ Now, using Stirling approximation
$$log(B)=frac1xleft(x (log (x)-1)+frac12 left(log (2 pi )+log
left(xright)right)+frac112
x+Oleft(frac1x^3right)right)$$
$$log(B)= (log (x)-1)+frac12x left(log (2 pi )+log
left(xright)right)+Oleft(frac1x^2right)$$ Continuing with Taylor
$$B=e^log(B)=fracxe+fraclog (2 pi )+log left(xright)2 e+Oleft(frac1xright)$$
$$A=fraclog (2 pi x)2 e+Oleft(frac1xright)tag 1$$ Pushing further the expansion of $B$, you would get
$$A=fraclog (2 pi x)2 e+frac3 log ^2left(2 pi xright)+224 e x+Oleft(frac1x^2right)tag 2$$ For illustration purposes, let $x=10^k$ and computing
$$left(
beginarraycccc
k & (1) & (2) & textexact \
1 & 0.761595453 & 0.843495044 & 0.849934276 \
2 & 1.185132311 & 1.204528536 & 1.204745228 \
3 & 1.608669170 & 1.612217034 & 1.612222301 \
4 & 2.032206029 & 2.032770401 & 2.032770506
endarray
right)$$
answered Aug 2 at 4:37
Claude Leibovici
111k1054126
(+1) So you have confirmed that the Maple expression is correct.
– gammatester
Aug 2 at 10:27
add a comment |Â
up vote
2
down vote
up vote
2
down vote
In the same spirit as in previous answers, considering $$A=(x!)^frac1x-fracxe=B-fracxe$$
$$B=(x!)^frac1ximplies log(B)=frac1xlog(x!)$$ Now, using Stirling approximation
$$log(B)=frac1xleft(x (log (x)-1)+frac12 left(log (2 pi )+log
left(xright)right)+frac112
x+Oleft(frac1x^3right)right)$$
$$log(B)= (log (x)-1)+frac12x left(log (2 pi )+log
left(xright)right)+Oleft(frac1x^2right)$$ Continuing with Taylor
$$B=e^log(B)=fracxe+fraclog (2 pi )+log left(xright)2 e+Oleft(frac1xright)$$
$$A=fraclog (2 pi x)2 e+Oleft(frac1xright)tag 1$$ Pushing further the expansion of $B$, you would get
$$A=fraclog (2 pi x)2 e+frac3 log ^2left(2 pi xright)+224 e x+Oleft(frac1x^2right)tag 2$$ For illustration purposes, let $x=10^k$ and computing
$$left(
beginarraycccc
k & (1) & (2) & textexact \
1 & 0.761595453 & 0.843495044 & 0.849934276 \
2 & 1.185132311 & 1.204528536 & 1.204745228 \
3 & 1.608669170 & 1.612217034 & 1.612222301 \
4 & 2.032206029 & 2.032770401 & 2.032770506
endarray
right)$$
answered Aug 2 at 4:37
Claude Leibovici
111k1054126
In the same spirit as in previous answers, considering $$A=(x!)^frac1x-fracxe=B-fracxe$$
$$B=(x!)^frac1ximplies log(B)=frac1xlog(x!)$$ Now, using Stirling approximation
$$log(B)=frac1xleft(x (log (x)-1)+frac12 left(log (2 pi )+log
left(xright)right)+frac112
x+Oleft(frac1x^3right)right)$$
$$log(B)= (log (x)-1)+frac12x left(log (2 pi )+log
left(xright)right)+Oleft(frac1x^2right)$$ Continuing with Taylor
$$B=e^log(B)=fracxe+fraclog (2 pi )+log left(xright)2 e+Oleft(frac1xright)$$
$$A=fraclog (2 pi x)2 e+Oleft(frac1xright)tag 1$$ Pushing further the expansion of $B$, you would get
$$A=fraclog (2 pi x)2 e+frac3 log ^2left(2 pi xright)+224 e x+Oleft(frac1x^2right)tag 2$$ For illustration purposes, let $x=10^k$ and computing
$$left(
beginarraycccc
k & (1) & (2) & textexact \
1 & 0.761595453 & 0.843495044 & 0.849934276 \
2 & 1.185132311 & 1.204528536 & 1.204745228 \
3 & 1.608669170 & 1.612217034 & 1.612222301 \
4 & 2.032206029 & 2.032770401 & 2.032770506
endarray
right)$$
answered Aug 2 at 4:37
Claude Leibovici
111k1054126
answered Aug 2 at 4:37
Claude Leibovici
111k1054126
answered Aug 2 at 4:37
Claude Leibovici
111k1054126
answered Aug 2 at 4:37
Claude Leibovici
111k1054126
111k1054126
(+1) So you have confirmed that the Maple expression is correct.
– gammatester
Aug 2 at 10:27
add a comment |Â
(+1) So you have confirmed that the Maple expression is correct.
– gammatester
Aug 2 at 10:27
(+1) So you have confirmed that the Maple expression is correct.
– gammatester
Aug 2 at 10:27
(+1) So you have confirmed that the Maple expression is correct.
– gammatester
Aug 2 at 10:27
add a comment |Â
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1
Have you tried Stirling's formula?
– Lord Shark the Unknown
Aug 1 at 17:32
If the limit is actually $infty$, the sequence diverges very slow. With PARI/GP, I got about value $3$ for $x=2000000$
– Peter
Aug 1 at 17:35
$log(n!)approx nlog n -n$
– saulspatz
Aug 1 at 17:41
1
If Maple and I did not make any mistake you have $$Gamma(x+1)^1/x-x/e = frac12eln(2pi x) + O(1/x), quad xrightarrow infty$$
– gammatester
Aug 1 at 17:44
@saulspatz unless I'm mistaken, your formula would imply that the limit does not blow up to infinity.
– Blake Steines
Aug 1 at 17:45