Let A_n =(3/4)-(3/4)^2+(3/4)^3-…..+ (-1)^n -1 (3/4)^n and B_n =1- A_n. Then, the least odd natural number p, so that B_n > A_n, for all n ≥ p, is:

Let $\displaystyle \mathrm{A}_{\mathrm{n}}=\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3-\ldots . .+$ $\displaystyle (-1)^{\mathrm{n}-1}\left(\frac{3}{4}\right)^{\mathrm{n}}$ and $\displaystyle \mathrm{B}_{\mathrm{n}}=1-\mathrm{A}_{\mathrm{n}}$. Then, the least odd natural number p, so that $\displaystyle \mathrm{B}_{\mathrm{n}}>\mathrm{A}_{\mathrm{n}}$, for all $\displaystyle \mathrm{n} \geqslant \mathrm{p}$, is :