Beta function $nB(frac{3}{4}, n-1) neq 1 $
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1
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I need to know if the equation $nB(frac{3}{4}, n-1) neq 1$ is true $forall n inmathbb{N}$. I do not idea whether it is true or not. I was trying to prove it but I have not idea how to deal with the $B(frac{3}{4}, n-1)$.
Thank you in advance.
real-analysis beta-function
asked Nov 23 at 17:36
Waney
906
add a comment |
up vote
1
down vote
favorite
I need to know if the equation $nB(frac{3}{4}, n-1) neq 1$ is true $forall n inmathbb{N}$. I do not idea whether it is true or not. I was trying to prove it but I have not idea how to deal with the $B(frac{3}{4}, n-1)$.
Thank you in advance.
real-analysis beta-function
asked Nov 23 at 17:36
Waney
906
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need to know if the equation $nB(frac{3}{4}, n-1) neq 1$ is true $forall n inmathbb{N}$. I do not idea whether it is true or not. I was trying to prove it but I have not idea how to deal with the $B(frac{3}{4}, n-1)$.
Thank you in advance.
real-analysis beta-function
asked Nov 23 at 17:36
Waney
906
I need to know if the equation $nB(frac{3}{4}, n-1) neq 1$ is true $forall n inmathbb{N}$. I do not idea whether it is true or not. I was trying to prove it but I have not idea how to deal with the $B(frac{3}{4}, n-1)$.
Thank you in advance.
real-analysis beta-function
real-analysis beta-function
asked Nov 23 at 17:36
Waney
906
asked Nov 23 at 17:36
Waney
906
edited Nov 24 at 10:41
asked Nov 23 at 17:36
Waney
906
asked Nov 23 at 17:36
Waney
906
asked Nov 23 at 17:36
Waney
906
906
add a comment |
add a comment |
1 Answer
1
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oldest
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up vote
3
down vote
accepted
Define $$u_n:=noperatorname{B}(tfrac{3}{4},,n-1)=frac{nGamma (tfrac{3}{4})Gamma(n-1)}{Gamma (n-tfrac{1}{4})}$$so that$$u_2=frac{2Gamma (tfrac{3}{4})Gamma (1)}{Gamma (tfrac{7}{4})}=frac{8}{3},,frac{u_{n+1}}{u_n}=frac{4(n^2-1)}{n(4n-1)}=1+frac{n-4}{n(4n-1)}.$$Thus$$u_3=frac{16}{7},,u_4=u_5=frac{512}{231},$$and thereafter the sequence is increasing.
answered Nov 23 at 17:48
J.G.
20.1k21932
I am quite confused about why the sequence is increasing, because $u_2$ is greater than $u_3$ and so on. Thanks
– Waney
Nov 24 at 10:49
Sorry, the sequence is not monotonic, so now it is clear to me, thanks again.
– Waney
Nov 24 at 11:35
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Define $$u_n:=noperatorname{B}(tfrac{3}{4},,n-1)=frac{nGamma (tfrac{3}{4})Gamma(n-1)}{Gamma (n-tfrac{1}{4})}$$so that$$u_2=frac{2Gamma (tfrac{3}{4})Gamma (1)}{Gamma (tfrac{7}{4})}=frac{8}{3},,frac{u_{n+1}}{u_n}=frac{4(n^2-1)}{n(4n-1)}=1+frac{n-4}{n(4n-1)}.$$Thus$$u_3=frac{16}{7},,u_4=u_5=frac{512}{231},$$and thereafter the sequence is increasing.
answered Nov 23 at 17:48
J.G.
20.1k21932
I am quite confused about why the sequence is increasing, because $u_2$ is greater than $u_3$ and so on. Thanks
– Waney
Nov 24 at 10:49
Sorry, the sequence is not monotonic, so now it is clear to me, thanks again.
– Waney
Nov 24 at 11:35
add a comment |
up vote
3
down vote
accepted
Define $$u_n:=noperatorname{B}(tfrac{3}{4},,n-1)=frac{nGamma (tfrac{3}{4})Gamma(n-1)}{Gamma (n-tfrac{1}{4})}$$so that$$u_2=frac{2Gamma (tfrac{3}{4})Gamma (1)}{Gamma (tfrac{7}{4})}=frac{8}{3},,frac{u_{n+1}}{u_n}=frac{4(n^2-1)}{n(4n-1)}=1+frac{n-4}{n(4n-1)}.$$Thus$$u_3=frac{16}{7},,u_4=u_5=frac{512}{231},$$and thereafter the sequence is increasing.
answered Nov 23 at 17:48
J.G.
20.1k21932
I am quite confused about why the sequence is increasing, because $u_2$ is greater than $u_3$ and so on. Thanks
– Waney
Nov 24 at 10:49
Sorry, the sequence is not monotonic, so now it is clear to me, thanks again.
– Waney
Nov 24 at 11:35
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Define $$u_n:=noperatorname{B}(tfrac{3}{4},,n-1)=frac{nGamma (tfrac{3}{4})Gamma(n-1)}{Gamma (n-tfrac{1}{4})}$$so that$$u_2=frac{2Gamma (tfrac{3}{4})Gamma (1)}{Gamma (tfrac{7}{4})}=frac{8}{3},,frac{u_{n+1}}{u_n}=frac{4(n^2-1)}{n(4n-1)}=1+frac{n-4}{n(4n-1)}.$$Thus$$u_3=frac{16}{7},,u_4=u_5=frac{512}{231},$$and thereafter the sequence is increasing.
answered Nov 23 at 17:48
J.G.
20.1k21932
Define $$u_n:=noperatorname{B}(tfrac{3}{4},,n-1)=frac{nGamma (tfrac{3}{4})Gamma(n-1)}{Gamma (n-tfrac{1}{4})}$$so that$$u_2=frac{2Gamma (tfrac{3}{4})Gamma (1)}{Gamma (tfrac{7}{4})}=frac{8}{3},,frac{u_{n+1}}{u_n}=frac{4(n^2-1)}{n(4n-1)}=1+frac{n-4}{n(4n-1)}.$$Thus$$u_3=frac{16}{7},,u_4=u_5=frac{512}{231},$$and thereafter the sequence is increasing.
answered Nov 23 at 17:48
J.G.
20.1k21932
answered Nov 23 at 17:48
J.G.
20.1k21932
answered Nov 23 at 17:48
J.G.
20.1k21932
answered Nov 23 at 17:48
J.G.
20.1k21932
20.1k21932
I am quite confused about why the sequence is increasing, because $u_2$ is greater than $u_3$ and so on. Thanks
– Waney
Nov 24 at 10:49
Sorry, the sequence is not monotonic, so now it is clear to me, thanks again.
– Waney
Nov 24 at 11:35
add a comment |
I am quite confused about why the sequence is increasing, because $u_2$ is greater than $u_3$ and so on. Thanks
– Waney
Nov 24 at 10:49
Sorry, the sequence is not monotonic, so now it is clear to me, thanks again.
– Waney
Nov 24 at 11:35
I am quite confused about why the sequence is increasing, because $u_2$ is greater than $u_3$ and so on. Thanks
– Waney
Nov 24 at 10:49
I am quite confused about why the sequence is increasing, because $u_2$ is greater than $u_3$ and so on. Thanks
– Waney
Nov 24 at 10:49
Sorry, the sequence is not monotonic, so now it is clear to me, thanks again.
– Waney
Nov 24 at 11:35
Sorry, the sequence is not monotonic, so now it is clear to me, thanks again.
– Waney
Nov 24 at 11:35
add a comment |
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