I want to prove that this sequence
$$A_n:=\frac{3^m \sqrt{\pi }\,\, \Gamma\left[\frac{1}{2}+m\right] \text{Hypergeometric2F1}\left[\frac{1}{2},1,1+m,\frac{3}{4}\right]}{2 \,\,\Gamma[m+1]}$$
is increasing $\forall n=1,2,3...$
I've tried with classical methods $A_{n}-A_{n+1}<0$ or $\displaystyle\frac{A_n}{A_{n+1}}<1$ but I'm not able.
Any help will be welcome.
I do not know how much this could help.
If $$a_m=\frac{ 3^m \sqrt{\pi }\,\,\Gamma \left(m+\frac{1}{2}\right)}{2 \Gamma (m+1)}\, _2F_1\left(\frac{1}{2},1;m+1;\frac{3}{4}\right)$$ then $$\frac{a_{m+1}}{a_m}=\frac{3(2m+1) }{2(m+1) }\frac{_2F_1\left(\frac{1}{2},1;m+2;\frac{3}{4}\right)} {_2F_1\left(\frac{1}{2},1;m+1;\frac{3}{4}\right)}$$
Now, in this page, the fourth recurrence identity for consecutive neighbors may be of interest. For large $m$, it seems that $$\frac{a_{m+1}}{a_m}\approx\frac{3(2m+1) }{2(m+1) }=3-\frac{3}{2 (m+1)}\approx 3$$
Edit
Just for the fun of playing with small numbers, I produced a table of $$b_m=\frac{_2F_1\left(\frac{1}{2},1;m+2;\frac{3}{4}\right)} {_2F_1\left(\frac{1}{2},1;m+1;\frac{3}{4}\right)}$$ $$\left( \begin{array}{cc} m & b_m \\ 1 & \frac{8}{9} \\ 2 & \frac{19}{20} \\ 3 & \frac{388}{399} \\ 4 & \frac{10295}{10476} \\ 5 & \frac{22382}{22649} \\ 6 & \frac{865445}{872898} \\ 7 & \frac{5527624}{5563575} \\ 8 & \frac{93494361}{93969608} \\ 9 & \frac{1769172910}{1776392859} \\ 10 & \frac{22217196067}{22291578666} \\ 11 & \frac{46324620476}{46454137231} \\ 12 & \frac{3466132816211}{3474346535700} \end{array} \right)$$