Beta function $nB(frac{3}{4}, n-1) neq 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.










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    up vote
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    down vote

    favorite
    1












    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.










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite
      1









      up vote
      1
      down vote

      favorite
      1






      1





      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.










      share|cite|improve this question















      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






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      edited Nov 24 at 10:41

























      asked Nov 23 at 17:36









      Waney

      906




      906






















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          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.






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          • 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











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          up vote
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          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.






          share|cite|improve this answer





















          • 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















          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.






          share|cite|improve this answer





















          • 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













          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 at 17:48









          J.G.

          21.5k21934




          21.5k21934












          • 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












          • 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


















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