Sum of the series $\sum_{k=1}^\infty ((\frac{x+1}{2})^2)^k$ Find the sum of the series for those values  of $$\sum_{k=1}^\infty ((\frac{x+1}{2})^2)^k$$
I have found that $$a=\frac{x^2+2x+1}{4}$$ and $$r=\frac{x^2+2x+1}{4}$$
The sum of the series would than equal $$=\frac{a}{1-r}=\frac{(\frac{x^2+2x+1}{4})}{1-(\frac{x^2+2x+1}{4})}=\frac{x^2+2x+1}{-x^2-2x+3}$$
I am now trying to figure out where this would be correct, but I am having trouble. I know that it has to be $|r|<1$ to be correct. I did
$$\frac{x^2+2x+1}{4}<1, which = x < 1$$, however, when I do the other side, I run into a problem
$$\frac{x^2+2x+1}{4}>-1$$ $$x^2+2x+1>-4$$ $$x^2+2x+5>0$$
This does not simplify anymore. Does that mean that it is correct for all x < 1 or is there something else I have to do?

Note that $r=\left(\frac{x+1}{2}\right)^2$. We have
$$\left(\frac{x+1}{2}\right)^2\lt 1$$
if and only if
$$-1\lt \frac{x+1}{2}\lt 1.$$
The right-hand inequality holds if and only if $x\lt 1$.
The left-hand inequality holds if and only if $-3\lt x$.
