Condition for infinite sum

For the given series to converge, $|2+4x| > 1$ $\implies \begin{cases} 2+4x > 1 & \implies x > - \frac 14 \\ 2+4x < -1 & \implies x < - \frac 34 \end{cases}$ . Therefore, $\left(\dfrac ab \right)^2 = \left(\dfrac {-\frac 34}{-\frac 14} \right)^2 = \boxed 9$ .

Assume $\dfrac 1{2+4x} + \dfrac 1{(2+4x)^2} + \dfrac 1{(2+4x)^3} + \ldots = S$ .

Here, first term, $u_1 = \dfrac 1{2+4x}$ and second term $u_2 = \dfrac 1{(2+4x)^2}$ .

So general ratio, $r = \dfrac{u_2}{u_1} = \dfrac{\frac 1{(2+4x)^2}}{\frac 1{2+4x}} = \dfrac 1{(2+4x)^2} \times (2+4x) = \dfrac 1{2+4x}.$

$S$ will only have a infinite sum if $-1 < r < 1$ . So $-1 < \dfrac 1{2+4x} < 1$ . Now,

$\begin{aligned} -1 & < \dfrac 1{2+4x} \\ \Rightarrow \dfrac 1{2+4x} & > -1 \\ \Rightarrow 2 + 4x & < -1 \\ \Rightarrow 4x & < -3 \\ \implies x & < -\dfrac 34 \\ \end{aligned}$

Again,

$\begin{aligned} \dfrac 1{2+4x} & < 1 \\ \Rightarrow 2+4x & > 1 \\ \Rightarrow 4x & > -1 \\ \implies x & > -\dfrac 14 \\ \end{aligned}$

So $a = -\dfrac 34$ and $b = -\dfrac 14$ . Hence $\left(\dfrac ab \right)^2 = \left(\dfrac{-\frac 34}{-\frac 14}\right)^2 = \left(\dfrac 34 \times 4 \right)^2 = 3^2 = \boxed 9$ .