Sequence and Series: JEE Main 2018 Mathematics question

Let    A_n = $\left( {{3 \over 4}} \right) - {\left( {{3 \over 4}} \right)^2} + {\left( {{3 \over 4}} \right)^3}$ $-$. . . . . + ($-$1)^n-1 ${\left( {{3 \over 4}} \right)^n}$    and    B_n = 1 $-$ A_n. Then, the least dd natural numbr p, so that B_n > A_n , for all n$\ge$ p, is :