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Why $lim_ntoinfty fracGammaleft(n - frac12right)Gammaleft(nright) = e^-frac12$
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Why is it true that
$$lim_ntoinfty fracGammaleft(n - frac12right)Gammaleft(nright) = e^-frac12$$
I only know the integral definition of gamma function. My notes writes
$$lim_ntoinfty fracGammaleft(n - frac12right)Gammaleft(nright) = lim_nprod_k=1^n-1left(1 - frac1/2kright) = e^-1/2$$
I don't know why the first equality holds, nor why the second equality holds...
gamma-function
asked Jul 28 at 3:30
3x89g2
3,8301932
add a comment |Â
up vote
0
down vote
favorite
Why is it true that
$$lim_ntoinfty fracGammaleft(n - frac12right)Gammaleft(nright) = e^-frac12$$
I only know the integral definition of gamma function. My notes writes
$$lim_ntoinfty fracGammaleft(n - frac12right)Gammaleft(nright) = lim_nprod_k=1^n-1left(1 - frac1/2kright) = e^-1/2$$
I don't know why the first equality holds, nor why the second equality holds...
gamma-function
asked Jul 28 at 3:30
3x89g2
3,8301932
1
The second equality definitely does not hold. $sum_k frac12k = +infty implies prod_k=1^infty (1-frac12k) = 0$
– mathworker21
Jul 28 at 3:34
1
Is it even true? W.A. seems to say otherwise (although I know WA isn't always correct): wolframalpha.com/input/?i=limit+of+gamma(n-0.5)%2Fgamma(n)
– Brenton
Jul 28 at 3:35
@Brenton I am not sure, to be honest. This note could be wrong. Anyway, I still need to evaluate the limit but not sure how to. My guess was that the limit would be $1$ (both $Gamma(n-1/2)$ and $Gamma(n)$ grows at similar rate?) but don't know if it's true or not.
– 3x89g2
Jul 28 at 3:38
As for why you should expect it to be zero: what is the limit of $(n - 1)! / n!$? (Then you can compare $ frac Gamma( n - 1/2 - 1/2) Gamma(n - 1/2) frac Gamma( n - 1/2) Gamma(n)= Gamma(n - 1)/Gamma(n) = (n-2)!/(n-1)!$, which you can turn into a proof, provided you can argue that the limit exists...)
– Lorenzo
Jul 28 at 3:53
To get some intuition, notice that $$fracGamma(n-k)Gamma(n)=frac1(n-1)(n-2)cdots(n-k)sim frac1n^kquadtextas ntoinfty$$ for each positive integer $k$. So we can expect that a similar behavior persists for non-integral $k$. Indeed, this extrapolation is justified by the Stirling's approximation.
– Sangchul Lee
Jul 28 at 4:05
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Why is it true that
$$lim_ntoinfty fracGammaleft(n - frac12right)Gammaleft(nright) = e^-frac12$$
I only know the integral definition of gamma function. My notes writes
$$lim_ntoinfty fracGammaleft(n - frac12right)Gammaleft(nright) = lim_nprod_k=1^n-1left(1 - frac1/2kright) = e^-1/2$$
I don't know why the first equality holds, nor why the second equality holds...
gamma-function
asked Jul 28 at 3:30
3x89g2
3,8301932
Why is it true that
$$lim_ntoinfty fracGammaleft(n - frac12right)Gammaleft(nright) = e^-frac12$$
I only know the integral definition of gamma function. My notes writes
$$lim_ntoinfty fracGammaleft(n - frac12right)Gammaleft(nright) = lim_nprod_k=1^n-1left(1 - frac1/2kright) = e^-1/2$$
I don't know why the first equality holds, nor why the second equality holds...
gamma-function
asked Jul 28 at 3:30
3x89g2
3,8301932
asked Jul 28 at 3:30
3x89g2
3,8301932
asked Jul 28 at 3:30
3x89g2
3,8301932
asked Jul 28 at 3:30
3x89g2
3,8301932
3,8301932
1
The second equality definitely does not hold. $sum_k frac12k = +infty implies prod_k=1^infty (1-frac12k) = 0$
– mathworker21
Jul 28 at 3:34
1
Is it even true? W.A. seems to say otherwise (although I know WA isn't always correct): wolframalpha.com/input/?i=limit+of+gamma(n-0.5)%2Fgamma(n)
– Brenton
Jul 28 at 3:35
@Brenton I am not sure, to be honest. This note could be wrong. Anyway, I still need to evaluate the limit but not sure how to. My guess was that the limit would be $1$ (both $Gamma(n-1/2)$ and $Gamma(n)$ grows at similar rate?) but don't know if it's true or not.
– 3x89g2
Jul 28 at 3:38
As for why you should expect it to be zero: what is the limit of $(n - 1)! / n!$? (Then you can compare $ frac Gamma( n - 1/2 - 1/2) Gamma(n - 1/2) frac Gamma( n - 1/2) Gamma(n)= Gamma(n - 1)/Gamma(n) = (n-2)!/(n-1)!$, which you can turn into a proof, provided you can argue that the limit exists...)
– Lorenzo
Jul 28 at 3:53
To get some intuition, notice that $$fracGamma(n-k)Gamma(n)=frac1(n-1)(n-2)cdots(n-k)sim frac1n^kquadtextas ntoinfty$$ for each positive integer $k$. So we can expect that a similar behavior persists for non-integral $k$. Indeed, this extrapolation is justified by the Stirling's approximation.
– Sangchul Lee
Jul 28 at 4:05
add a comment |Â
1
The second equality definitely does not hold. $sum_k frac12k = +infty implies prod_k=1^infty (1-frac12k) = 0$
– mathworker21
Jul 28 at 3:34
1
Is it even true? W.A. seems to say otherwise (although I know WA isn't always correct): wolframalpha.com/input/?i=limit+of+gamma(n-0.5)%2Fgamma(n)
– Brenton
Jul 28 at 3:35
@Brenton I am not sure, to be honest. This note could be wrong. Anyway, I still need to evaluate the limit but not sure how to. My guess was that the limit would be $1$ (both $Gamma(n-1/2)$ and $Gamma(n)$ grows at similar rate?) but don't know if it's true or not.
– 3x89g2
Jul 28 at 3:38
As for why you should expect it to be zero: what is the limit of $(n - 1)! / n!$? (Then you can compare $ frac Gamma( n - 1/2 - 1/2) Gamma(n - 1/2) frac Gamma( n - 1/2) Gamma(n)= Gamma(n - 1)/Gamma(n) = (n-2)!/(n-1)!$, which you can turn into a proof, provided you can argue that the limit exists...)
– Lorenzo
Jul 28 at 3:53
To get some intuition, notice that $$fracGamma(n-k)Gamma(n)=frac1(n-1)(n-2)cdots(n-k)sim frac1n^kquadtextas ntoinfty$$ for each positive integer $k$. So we can expect that a similar behavior persists for non-integral $k$. Indeed, this extrapolation is justified by the Stirling's approximation.
– Sangchul Lee
Jul 28 at 4:05
1
1
The second equality definitely does not hold. $sum_k frac12k = +infty implies prod_k=1^infty (1-frac12k) = 0$
– mathworker21
Jul 28 at 3:34
The second equality definitely does not hold. $sum_k frac12k = +infty implies prod_k=1^infty (1-frac12k) = 0$
– mathworker21
Jul 28 at 3:34
1
1
Is it even true? W.A. seems to say otherwise (although I know WA isn't always correct): wolframalpha.com/input/?i=limit+of+gamma(n-0.5)%2Fgamma(n)
– Brenton
Jul 28 at 3:35
Is it even true? W.A. seems to say otherwise (although I know WA isn't always correct): wolframalpha.com/input/?i=limit+of+gamma(n-0.5)%2Fgamma(n)
– Brenton
Jul 28 at 3:35
@Brenton I am not sure, to be honest. This note could be wrong. Anyway, I still need to evaluate the limit but not sure how to. My guess was that the limit would be $1$ (both $Gamma(n-1/2)$ and $Gamma(n)$ grows at similar rate?) but don't know if it's true or not.
– 3x89g2
Jul 28 at 3:38
@Brenton I am not sure, to be honest. This note could be wrong. Anyway, I still need to evaluate the limit but not sure how to. My guess was that the limit would be $1$ (both $Gamma(n-1/2)$ and $Gamma(n)$ grows at similar rate?) but don't know if it's true or not.
– 3x89g2
Jul 28 at 3:38
As for why you should expect it to be zero: what is the limit of $(n - 1)! / n!$? (Then you can compare $ frac Gamma( n - 1/2 - 1/2) Gamma(n - 1/2) frac Gamma( n - 1/2) Gamma(n)= Gamma(n - 1)/Gamma(n) = (n-2)!/(n-1)!$, which you can turn into a proof, provided you can argue that the limit exists...)
– Lorenzo
Jul 28 at 3:53
As for why you should expect it to be zero: what is the limit of $(n - 1)! / n!$? (Then you can compare $ frac Gamma( n - 1/2 - 1/2) Gamma(n - 1/2) frac Gamma( n - 1/2) Gamma(n)= Gamma(n - 1)/Gamma(n) = (n-2)!/(n-1)!$, which you can turn into a proof, provided you can argue that the limit exists...)
– Lorenzo
Jul 28 at 3:53
To get some intuition, notice that $$fracGamma(n-k)Gamma(n)=frac1(n-1)(n-2)cdots(n-k)sim frac1n^kquadtextas ntoinfty$$ for each positive integer $k$. So we can expect that a similar behavior persists for non-integral $k$. Indeed, this extrapolation is justified by the Stirling's approximation.
– Sangchul Lee
Jul 28 at 4:05
To get some intuition, notice that $$fracGamma(n-k)Gamma(n)=frac1(n-1)(n-2)cdots(n-k)sim frac1n^kquadtextas ntoinfty$$ for each positive integer $k$. So we can expect that a similar behavior persists for non-integral $k$. Indeed, this extrapolation is justified by the Stirling's approximation.
– Sangchul Lee
Jul 28 at 4:05
add a comment |Â
3 Answers
3
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up vote
5
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This is not true. If it were true, we'd have
begineqnarray*
lim_ntoinftyfracGammaleft(n-1right)Gamma(n)
&=&
lim_ntoinftyleft(fracGammaleft(n-frac12right)Gamma(n)cdotfracGamma(n-1)Gammaleft(n-frac12right)right)
\
&=&
lim_ntoinftyfracGammaleft(n-frac12right)Gamma(n)cdotlim_ntoinftyfracGamma(n-1)Gammaleft(n-frac12right)
\
&=&
mathrm e^-frac12cdotmathrm e^-frac12
\
&=&
mathrm e^-1;,
endeqnarray*
whereas this limit is in fact $0$ (since the quotient is $frac1n-1$).
The correct statement that this might be intended to state might be
$$
fracGammaleft(n-frac12right)Gamma(n)sim n^-frac12;.
$$
answered Jul 28 at 4:09
joriki
164k10179328
May I ask how do you get the third equal sign (How to evaluate those two limits)? If it's approximately $sqrtfrac1n$ then wouldn't it goes to zero as $ntoinfty$?
– 3x89g2
Aug 2 at 3:32
@3x89g2: Indeed. It's a proof by contradiction. It assumes the OP's false claim in order to disprove it by deriving.a contradiction from it. It's prefaced with: "This is not true. If it were true, we'd have ...". The third equals sign uses the OP's false claim twice (once directly, once shifted).
– joriki
Aug 2 at 5:00
Ah right... I misread. Thank you.
– 3x89g2
Aug 2 at 5:01
add a comment |Â
up vote
1
down vote
Considering the general case$$f_a=fracGamma (n-a)Gamma (n)implies log(f_a)=log (Gamma (n-a))-log (Gamma (n))$$ and use Stirling approximation to get
$$log(f_a)=-a log left(nright)+fraca(a+1)2
n+Oleft(frac1n^2right)$$ Continuing with Taylor series
$$f_a=e^log(f_a)=n^-a left(1+fraca(a+1)2
n+Oleft(frac1n^2right) right)$$
answered Jul 28 at 5:06
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
The first equality is not true reads the comments.
for the second equality cannot be true
note that by telescopic sum we have
$$ lnleft(prod_k=1^n-1left(1 - frac1/2kright)right) = sum_k=1^n-1lnleft(2k - 1right)-lnleft(2k right)= sum_j=2^2n-2lnleft(j- 1right)-lnleft(j right)\ = lnleft(2- 1right)-lnleft(2n-2 right)to-infty $$
Whence
$$prod_k=1^n-1left(1 - frac1/2kright) to0.$$
answered Jul 28 at 4:35
Guy Fsone
16.8k42671
Your summation index change isn't valid. In $$ sum_k=2^2n-2lnleft(j- 1right)-lnleft(j right)$$ you probably meant the summation index to be $j$ rather than $k$; but that sum isn't equal to the previous one; you now have twice as many terms.
– joriki
Jul 28 at 5:50
Also, the first equation is not true. $Gammaleft(n-frac12right)$ isn't a product of $n$ factors of the form $k-frac12$, as appears to be assumed in that equation.
– joriki
Jul 28 at 5:53
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
accepted
This is not true. If it were true, we'd have
begineqnarray*
lim_ntoinftyfracGammaleft(n-1right)Gamma(n)
&=&
lim_ntoinftyleft(fracGammaleft(n-frac12right)Gamma(n)cdotfracGamma(n-1)Gammaleft(n-frac12right)right)
\
&=&
lim_ntoinftyfracGammaleft(n-frac12right)Gamma(n)cdotlim_ntoinftyfracGamma(n-1)Gammaleft(n-frac12right)
\
&=&
mathrm e^-frac12cdotmathrm e^-frac12
\
&=&
mathrm e^-1;,
endeqnarray*
whereas this limit is in fact $0$ (since the quotient is $frac1n-1$).
The correct statement that this might be intended to state might be
$$
fracGammaleft(n-frac12right)Gamma(n)sim n^-frac12;.
$$
answered Jul 28 at 4:09
joriki
164k10179328
May I ask how do you get the third equal sign (How to evaluate those two limits)? If it's approximately $sqrtfrac1n$ then wouldn't it goes to zero as $ntoinfty$?
– 3x89g2
Aug 2 at 3:32
@3x89g2: Indeed. It's a proof by contradiction. It assumes the OP's false claim in order to disprove it by deriving.a contradiction from it. It's prefaced with: "This is not true. If it were true, we'd have ...". The third equals sign uses the OP's false claim twice (once directly, once shifted).
– joriki
Aug 2 at 5:00
Ah right... I misread. Thank you.
– 3x89g2
Aug 2 at 5:01
add a comment |Â
up vote
5
down vote
accepted
This is not true. If it were true, we'd have
begineqnarray*
lim_ntoinftyfracGammaleft(n-1right)Gamma(n)
&=&
lim_ntoinftyleft(fracGammaleft(n-frac12right)Gamma(n)cdotfracGamma(n-1)Gammaleft(n-frac12right)right)
\
&=&
lim_ntoinftyfracGammaleft(n-frac12right)Gamma(n)cdotlim_ntoinftyfracGamma(n-1)Gammaleft(n-frac12right)
\
&=&
mathrm e^-frac12cdotmathrm e^-frac12
\
&=&
mathrm e^-1;,
endeqnarray*
whereas this limit is in fact $0$ (since the quotient is $frac1n-1$).
The correct statement that this might be intended to state might be
$$
fracGammaleft(n-frac12right)Gamma(n)sim n^-frac12;.
$$
answered Jul 28 at 4:09
joriki
164k10179328
May I ask how do you get the third equal sign (How to evaluate those two limits)? If it's approximately $sqrtfrac1n$ then wouldn't it goes to zero as $ntoinfty$?
– 3x89g2
Aug 2 at 3:32
@3x89g2: Indeed. It's a proof by contradiction. It assumes the OP's false claim in order to disprove it by deriving.a contradiction from it. It's prefaced with: "This is not true. If it were true, we'd have ...". The third equals sign uses the OP's false claim twice (once directly, once shifted).
– joriki
Aug 2 at 5:00
Ah right... I misread. Thank you.
– 3x89g2
Aug 2 at 5:01
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
This is not true. If it were true, we'd have
begineqnarray*
lim_ntoinftyfracGammaleft(n-1right)Gamma(n)
&=&
lim_ntoinftyleft(fracGammaleft(n-frac12right)Gamma(n)cdotfracGamma(n-1)Gammaleft(n-frac12right)right)
\
&=&
lim_ntoinftyfracGammaleft(n-frac12right)Gamma(n)cdotlim_ntoinftyfracGamma(n-1)Gammaleft(n-frac12right)
\
&=&
mathrm e^-frac12cdotmathrm e^-frac12
\
&=&
mathrm e^-1;,
endeqnarray*
whereas this limit is in fact $0$ (since the quotient is $frac1n-1$).
The correct statement that this might be intended to state might be
$$
fracGammaleft(n-frac12right)Gamma(n)sim n^-frac12;.
$$
answered Jul 28 at 4:09
joriki
164k10179328
This is not true. If it were true, we'd have
begineqnarray*
lim_ntoinftyfracGammaleft(n-1right)Gamma(n)
&=&
lim_ntoinftyleft(fracGammaleft(n-frac12right)Gamma(n)cdotfracGamma(n-1)Gammaleft(n-frac12right)right)
\
&=&
lim_ntoinftyfracGammaleft(n-frac12right)Gamma(n)cdotlim_ntoinftyfracGamma(n-1)Gammaleft(n-frac12right)
\
&=&
mathrm e^-frac12cdotmathrm e^-frac12
\
&=&
mathrm e^-1;,
endeqnarray*
whereas this limit is in fact $0$ (since the quotient is $frac1n-1$).
The correct statement that this might be intended to state might be
$$
fracGammaleft(n-frac12right)Gamma(n)sim n^-frac12;.
$$
answered Jul 28 at 4:09
joriki
164k10179328
answered Jul 28 at 4:09
joriki
164k10179328
answered Jul 28 at 4:09
joriki
164k10179328
answered Jul 28 at 4:09
joriki
164k10179328
164k10179328
May I ask how do you get the third equal sign (How to evaluate those two limits)? If it's approximately $sqrtfrac1n$ then wouldn't it goes to zero as $ntoinfty$?
– 3x89g2
Aug 2 at 3:32
@3x89g2: Indeed. It's a proof by contradiction. It assumes the OP's false claim in order to disprove it by deriving.a contradiction from it. It's prefaced with: "This is not true. If it were true, we'd have ...". The third equals sign uses the OP's false claim twice (once directly, once shifted).
– joriki
Aug 2 at 5:00
Ah right... I misread. Thank you.
– 3x89g2
Aug 2 at 5:01
add a comment |Â
May I ask how do you get the third equal sign (How to evaluate those two limits)? If it's approximately $sqrtfrac1n$ then wouldn't it goes to zero as $ntoinfty$?
– 3x89g2
Aug 2 at 3:32
@3x89g2: Indeed. It's a proof by contradiction. It assumes the OP's false claim in order to disprove it by deriving.a contradiction from it. It's prefaced with: "This is not true. If it were true, we'd have ...". The third equals sign uses the OP's false claim twice (once directly, once shifted).
– joriki
Aug 2 at 5:00
Ah right... I misread. Thank you.
– 3x89g2
Aug 2 at 5:01
May I ask how do you get the third equal sign (How to evaluate those two limits)? If it's approximately $sqrtfrac1n$ then wouldn't it goes to zero as $ntoinfty$?
– 3x89g2
Aug 2 at 3:32
May I ask how do you get the third equal sign (How to evaluate those two limits)? If it's approximately $sqrtfrac1n$ then wouldn't it goes to zero as $ntoinfty$?
– 3x89g2
Aug 2 at 3:32
@3x89g2: Indeed. It's a proof by contradiction. It assumes the OP's false claim in order to disprove it by deriving.a contradiction from it. It's prefaced with: "This is not true. If it were true, we'd have ...". The third equals sign uses the OP's false claim twice (once directly, once shifted).
– joriki
Aug 2 at 5:00
@3x89g2: Indeed. It's a proof by contradiction. It assumes the OP's false claim in order to disprove it by deriving.a contradiction from it. It's prefaced with: "This is not true. If it were true, we'd have ...". The third equals sign uses the OP's false claim twice (once directly, once shifted).
– joriki
Aug 2 at 5:00
Ah right... I misread. Thank you.
– 3x89g2
Aug 2 at 5:01
Ah right... I misread. Thank you.
– 3x89g2
Aug 2 at 5:01
add a comment |Â
up vote
1
down vote
Considering the general case$$f_a=fracGamma (n-a)Gamma (n)implies log(f_a)=log (Gamma (n-a))-log (Gamma (n))$$ and use Stirling approximation to get
$$log(f_a)=-a log left(nright)+fraca(a+1)2
n+Oleft(frac1n^2right)$$ Continuing with Taylor series
$$f_a=e^log(f_a)=n^-a left(1+fraca(a+1)2
n+Oleft(frac1n^2right) right)$$
answered Jul 28 at 5:06
Claude Leibovici
111k1055126
add a comment |Â
up vote
1
down vote
Considering the general case$$f_a=fracGamma (n-a)Gamma (n)implies log(f_a)=log (Gamma (n-a))-log (Gamma (n))$$ and use Stirling approximation to get
$$log(f_a)=-a log left(nright)+fraca(a+1)2
n+Oleft(frac1n^2right)$$ Continuing with Taylor series
$$f_a=e^log(f_a)=n^-a left(1+fraca(a+1)2
n+Oleft(frac1n^2right) right)$$
answered Jul 28 at 5:06
Claude Leibovici
111k1055126
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Considering the general case$$f_a=fracGamma (n-a)Gamma (n)implies log(f_a)=log (Gamma (n-a))-log (Gamma (n))$$ and use Stirling approximation to get
$$log(f_a)=-a log left(nright)+fraca(a+1)2
n+Oleft(frac1n^2right)$$ Continuing with Taylor series
$$f_a=e^log(f_a)=n^-a left(1+fraca(a+1)2
n+Oleft(frac1n^2right) right)$$
answered Jul 28 at 5:06
Claude Leibovici
111k1055126
Considering the general case$$f_a=fracGamma (n-a)Gamma (n)implies log(f_a)=log (Gamma (n-a))-log (Gamma (n))$$ and use Stirling approximation to get
$$log(f_a)=-a log left(nright)+fraca(a+1)2
n+Oleft(frac1n^2right)$$ Continuing with Taylor series
$$f_a=e^log(f_a)=n^-a left(1+fraca(a+1)2
n+Oleft(frac1n^2right) right)$$
answered Jul 28 at 5:06
Claude Leibovici
111k1055126
answered Jul 28 at 5:06
Claude Leibovici
111k1055126
answered Jul 28 at 5:06
Claude Leibovici
111k1055126
answered Jul 28 at 5:06
Claude Leibovici
111k1055126
111k1055126
add a comment |Â
add a comment |Â
up vote
0
down vote
The first equality is not true reads the comments.
for the second equality cannot be true
note that by telescopic sum we have
$$ lnleft(prod_k=1^n-1left(1 - frac1/2kright)right) = sum_k=1^n-1lnleft(2k - 1right)-lnleft(2k right)= sum_j=2^2n-2lnleft(j- 1right)-lnleft(j right)\ = lnleft(2- 1right)-lnleft(2n-2 right)to-infty $$
Whence
$$prod_k=1^n-1left(1 - frac1/2kright) to0.$$
answered Jul 28 at 4:35
Guy Fsone
16.8k42671
Your summation index change isn't valid. In $$ sum_k=2^2n-2lnleft(j- 1right)-lnleft(j right)$$ you probably meant the summation index to be $j$ rather than $k$; but that sum isn't equal to the previous one; you now have twice as many terms.
– joriki
Jul 28 at 5:50
Also, the first equation is not true. $Gammaleft(n-frac12right)$ isn't a product of $n$ factors of the form $k-frac12$, as appears to be assumed in that equation.
– joriki
Jul 28 at 5:53
add a comment |Â
up vote
0
down vote
The first equality is not true reads the comments.
for the second equality cannot be true
note that by telescopic sum we have
$$ lnleft(prod_k=1^n-1left(1 - frac1/2kright)right) = sum_k=1^n-1lnleft(2k - 1right)-lnleft(2k right)= sum_j=2^2n-2lnleft(j- 1right)-lnleft(j right)\ = lnleft(2- 1right)-lnleft(2n-2 right)to-infty $$
Whence
$$prod_k=1^n-1left(1 - frac1/2kright) to0.$$
answered Jul 28 at 4:35
Guy Fsone
16.8k42671
Your summation index change isn't valid. In $$ sum_k=2^2n-2lnleft(j- 1right)-lnleft(j right)$$ you probably meant the summation index to be $j$ rather than $k$; but that sum isn't equal to the previous one; you now have twice as many terms.
– joriki
Jul 28 at 5:50
Also, the first equation is not true. $Gammaleft(n-frac12right)$ isn't a product of $n$ factors of the form $k-frac12$, as appears to be assumed in that equation.
– joriki
Jul 28 at 5:53
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The first equality is not true reads the comments.
for the second equality cannot be true
note that by telescopic sum we have
$$ lnleft(prod_k=1^n-1left(1 - frac1/2kright)right) = sum_k=1^n-1lnleft(2k - 1right)-lnleft(2k right)= sum_j=2^2n-2lnleft(j- 1right)-lnleft(j right)\ = lnleft(2- 1right)-lnleft(2n-2 right)to-infty $$
Whence
$$prod_k=1^n-1left(1 - frac1/2kright) to0.$$
answered Jul 28 at 4:35
Guy Fsone
16.8k42671
The first equality is not true reads the comments.
for the second equality cannot be true
note that by telescopic sum we have
$$ lnleft(prod_k=1^n-1left(1 - frac1/2kright)right) = sum_k=1^n-1lnleft(2k - 1right)-lnleft(2k right)= sum_j=2^2n-2lnleft(j- 1right)-lnleft(j right)\ = lnleft(2- 1right)-lnleft(2n-2 right)to-infty $$
Whence
$$prod_k=1^n-1left(1 - frac1/2kright) to0.$$
answered Jul 28 at 4:35
Guy Fsone
16.8k42671
edited Jul 28 at 6:46
answered Jul 28 at 4:35
Guy Fsone
16.8k42671
answered Jul 28 at 4:35
Guy Fsone
16.8k42671
answered Jul 28 at 4:35
Guy Fsone
16.8k42671
16.8k42671
Your summation index change isn't valid. In $$ sum_k=2^2n-2lnleft(j- 1right)-lnleft(j right)$$ you probably meant the summation index to be $j$ rather than $k$; but that sum isn't equal to the previous one; you now have twice as many terms.
– joriki
Jul 28 at 5:50
Also, the first equation is not true. $Gammaleft(n-frac12right)$ isn't a product of $n$ factors of the form $k-frac12$, as appears to be assumed in that equation.
– joriki
Jul 28 at 5:53
add a comment |Â
Your summation index change isn't valid. In $$ sum_k=2^2n-2lnleft(j- 1right)-lnleft(j right)$$ you probably meant the summation index to be $j$ rather than $k$; but that sum isn't equal to the previous one; you now have twice as many terms.
– joriki
Jul 28 at 5:50
Also, the first equation is not true. $Gammaleft(n-frac12right)$ isn't a product of $n$ factors of the form $k-frac12$, as appears to be assumed in that equation.
– joriki
Jul 28 at 5:53
Your summation index change isn't valid. In $$ sum_k=2^2n-2lnleft(j- 1right)-lnleft(j right)$$ you probably meant the summation index to be $j$ rather than $k$; but that sum isn't equal to the previous one; you now have twice as many terms.
– joriki
Jul 28 at 5:50
Your summation index change isn't valid. In $$ sum_k=2^2n-2lnleft(j- 1right)-lnleft(j right)$$ you probably meant the summation index to be $j$ rather than $k$; but that sum isn't equal to the previous one; you now have twice as many terms.
– joriki
Jul 28 at 5:50
Also, the first equation is not true. $Gammaleft(n-frac12right)$ isn't a product of $n$ factors of the form $k-frac12$, as appears to be assumed in that equation.
– joriki
Jul 28 at 5:53
Also, the first equation is not true. $Gammaleft(n-frac12right)$ isn't a product of $n$ factors of the form $k-frac12$, as appears to be assumed in that equation.
– joriki
Jul 28 at 5:53
add a comment |Â
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1
The second equality definitely does not hold. $sum_k frac12k = +infty implies prod_k=1^infty (1-frac12k) = 0$
– mathworker21
Jul 28 at 3:34
1
Is it even true? W.A. seems to say otherwise (although I know WA isn't always correct): wolframalpha.com/input/?i=limit+of+gamma(n-0.5)%2Fgamma(n)
– Brenton
Jul 28 at 3:35
@Brenton I am not sure, to be honest. This note could be wrong. Anyway, I still need to evaluate the limit but not sure how to. My guess was that the limit would be $1$ (both $Gamma(n-1/2)$ and $Gamma(n)$ grows at similar rate?) but don't know if it's true or not.
– 3x89g2
Jul 28 at 3:38
As for why you should expect it to be zero: what is the limit of $(n - 1)! / n!$? (Then you can compare $ frac Gamma( n - 1/2 - 1/2) Gamma(n - 1/2) frac Gamma( n - 1/2) Gamma(n)= Gamma(n - 1)/Gamma(n) = (n-2)!/(n-1)!$, which you can turn into a proof, provided you can argue that the limit exists...)
– Lorenzo
Jul 28 at 3:53
To get some intuition, notice that $$fracGamma(n-k)Gamma(n)=frac1(n-1)(n-2)cdots(n-k)sim frac1n^kquadtextas ntoinfty$$ for each positive integer $k$. So we can expect that a similar behavior persists for non-integral $k$. Indeed, this extrapolation is justified by the Stirling's approximation.
– Sangchul Lee
Jul 28 at 4:05