$$A_n=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+....+\left(-1\right)^{n-1}\left(\frac{3}{4}\right)^n$$ $$B_n=1-A_n$$ Find the minimum natural number $n_0$ such that $B_n>A_n\quad\forall\quad n>n_0$
My attempt is as follows:
$$A_n=\frac{\frac{3}{4}\left(1-\left(\frac{-3}{4}\right)^n\right)}{1-\left(\frac{-3}{4}\right)}$$
$$A_n=\frac{3}{7}\left(1-\left(-1\right)^n\left(\frac{3}{4}\right)^n\right)$$
As we have to find minimum natural number $n_0$ such that $B_n>A_n\quad\forall\quad n>n_0$
$$1-A_n>A_n$$ $$1>2A_n$$ $$\frac{1}{2}>\frac{3}{7}\left(1-\left(-1\right)^n\left(\frac{3}{4}\right)^n\right)$$ $$\frac{7}{6}>\left(1-\left(-1\right)^n\left(\frac{3}{4}\right)^n\right)$$ $$\frac{1}{6}>\left(-1\right)^{n+1}\left(\frac{3}{4}\right)^n$$
It is easy to notice that if n is even, then this inequality always satisfies.
If n is odd, then we can write as:
$$2^{2n-1}>3^{n+1}$$
Lets check for odd values of n:
$n=1$
$$2>9$$
$n=3$
$${32>81}$$
$n=5$
$${512>729}$$
$n=7$
$${8192>6561}$$
So from $n>5$ , this equation will always be satisfied as 6 is even and for 7 we just checked.
So $n_0$ should be equal to $5$ but everywhere answer to this question is said to be 6.