$\frac34-(\frac34)^2+(\frac34)^3-...+(-1)^{n-1}(\frac34)^n$ 
Let $A_n=\frac34-(\frac34)^2+(\frac34)^3-...+(-1)^{n-1}(\frac34)^n$ and $B_n=1-A_n$, then find the least value of $n_0, n_0\in N$ such that $B_n>A_n,\forall n\ge n_0$.

My attempt:$A_n$ is a Geometric Progression. So, $$A_n=\frac37\left[1-\left(-\frac34\right)^n\right]$$Also,$$1-A_n>A_n$$
So,$$A_n<\frac12$$
So,$$1-\left(-\frac34\right)^n<\frac76$$
So, $$\left(-\frac34\right)^n>-\frac16$$
Putting values of $n$ as $1,2,3...$, but nothing concrete is appearing.
 A: If you are correct to this point, then you could proceed as follows:
If $n$ is odd then $n=2m+1$
$$\left(-\frac{3}{4}\right)^{2m+1}>-\frac16$$
So $\left(-\frac34\right )^{2m}<\frac{1}{6}\cdot \frac43$, where we multiplyed the inequality with $-\frac43$. Keep in mind that this changes the direction of the relation.
So we get:
$$\left(\left(-\frac34\right)^{2}\right)^m=\left(\frac{9}{16}\right)^m<\frac{4}{18}$$
Now take the logarithm and solve for $m$. 
$$m\log\left(\frac9{16}\right)<\log\left(\frac{2}{9}\right)\Leftrightarrow m>\frac{\log\left(\frac{2}{9}\right)}{\log\left(\frac{9}{16}\right)}$$
Again $\log(x)$ is negativ for $0<x<1$ so it changes the sign.
If $n$ is even, then the LHS is positive and the RHS is negative, so always true.
A: The negative terms in the sequence $u_n:=\left(-\frac34\right)^n$ aren't $>-\frac16$ until their moduli are $<\frac16$, which happens once $n$ is an odd number $\ge\frac{\ln 6}{\ln\frac43}\approx 6.23$, so take $n_0=6$.
A: $$(-\frac34)^n>-\frac16 $$
is satisfied for $n=6$ because for $n=6$ it is obvious and  $$(-\frac34)^7=-.1334..> -1/6$$
And the inequality stays valid for $n\ge 6$ because the sequence terms $(-3/4)^n$ are either positive or decreasing in absolute value.  
