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

Question

I need to know if the equation $nB(\frac{3}{4}, n-1) \neq 1$ is true $\forall n \in\mathbb{N}$. 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.

Answer 1

Define $$u_n:=n\operatorname{B}(\tfrac{3}{4},\,n-1)=\frac{n\Gamma (\tfrac{3}{4})\Gamma(n-1)}{\Gamma (n-\tfrac{1}{4})}$$so that$$u_2=\frac{2\Gamma (\tfrac{3}{4})\Gamma (1)}{\Gamma (\tfrac{7}{4})}=\frac{8}{3},\,\frac{u_{n+1}}{u_n}=\frac{4(n^2-1)}{n(4n-1)}=1+\frac{n-4}{n(4n-1)}.$$Thus$$u_3=\frac{16}{7},\,u_4=u_5=\frac{512}{231},$$and thereafter the sequence is increasing.