Let A_n = 34-34^2+34^3-+(-1)^n-134^n and B_n=1-A_n.

Question

Let $A_n = \dfrac34-\dfrac{3}{4^2}+\dfrac{3}{4^3}-\cdots+(-1)^{n-1}\dfrac{3}{4^n}$ and $B_n=1-A_n$. Then the least odd natural number $p$, so that $B_n>A_n$ for all $n\ge p$, is:

Accepted Answer

$A_n$ is a G.P. sum with $a= frac34$, $r=- frac34$: $A_n = \dfrac37\left[1-\left(-\dfrac34 ight)^n ight]$. $B_n>A_n \Leftrightarrow A_n<\dfrac12$: $\dfrac37\left[1-\left(-\dfrac34 ight)^n ight]<\dfrac12 \Rightarrow \left(-\dfrac34 ight)^n > -\dfrac16.$ For odd $n$, $\left(-\dfrac34 ight)^n = -\left(\dfrac34 ight)^n$, so $\left(\dfrac34 ight)^n<\dfrac16 \Rightarrow n>\dfrac{\ln6}{\ln(4/3)}\approx6.23.$ Least odd $n$: $p=7$.

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