Let $A_n=\left(\frac{3}{4}\right)-{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{4}\right)}^3\dots ..+(-1{)}^{n-1}{\left(\frac{3}{4}\right)}^n$and $B_n=1-A_n.$Then, the least odd natural number $p$, so that $B_n>A_n\text{, }$for all $n\ge p\text{, }$is

We have,

$A_n=\frac{3}{4}-{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{4}\right)}^3\dots \dots +(-1{)}^{n-1}{\left(\frac{3}{4}\right)}^n$

which is a G.P. with $a=\frac{3}{4},r=-\frac{3}{4}$

$\therefore A_n=\frac{3}{4}\left(\frac{1-{\left(-\frac{3}{4}\right)}^n}{1+\frac{3}{4}}\right)=\frac{3}{7}\left(1-{\left(\frac{-3}{4}\right)}^n\right)$

Also, $B_n=1-A_n$

Consider $B_n>A_n\Rightarrow 1-A_n>A_n$

$\begin{array}{l}\Rightarrow 2A_n<1\Rightarrow \frac{6}{7}\left(1-{\left(\frac{-3}{4}\right)}^n\right)<1\\ \Rightarrow 1-{\left(\frac{-3}{4}\right)}^n<\frac{7}{6}\Rightarrow (-1{)}^{n+1}{\left(\frac{3}{4}\right)}^n<\frac{1}{6}=\frac{1}{2\times 3}\\ \Rightarrow (-1{)}^{n+1}{3}^{n+1}<2^{2n-1}\Rightarrow (-3{)}^{n+1}<2^{2n-1}\end{array}$

which is true for all even natural numbers and for all 

odd natural numbers $\ge 7$ 

$\therefore$ Least odd natural number$p$ is 7.