Question

If $a_n=\frac{3}{4}-(\frac{3}{4}{)}^2+(\frac{3}{4}{)}^3+...+(-1{)}^{n-1}(\frac{3}{4}{)}^n$ and $b_n=1-a_n$ then find the minimum natural number $n_0$ such that $b_n>a_n\forall n>n_0$

Answer 1

$b_n=1-a_n>a_n\Rightarrow a_n<\frac{1}{2}$
Now $a_n=\frac{\frac{3}{4}(1-{(-\frac{3}{4})}^n)}{1+\frac{3}{4}}<\frac{1}{2}$.......(Sum of n terms of G.P. with common ratio $\frac{-3}{4}$)

$\Rightarrow {\left(\frac{-3}{4}\right)}^n>-\frac{1}{6}$

Obviously, it is true for all odd values of n.
But for
$n=1,-\frac{3}{4}<-\frac{1}{6}$
$n=3,{\left(\frac{-3}{4}\right)}^3=-\frac{27}{64}<-\frac{1}{6}$
$n=5,{(-\frac{3}{4})}^5=-\frac{243}{1024}<-\frac{1}{6}$
now for $n=7,{(-\frac{3}{4})}^7=-\frac{2187}{12288}>-\frac{1}{6}$
Therefore minimum $n_0=7$