Interesting Series

Let the sum be $S$ .

$\begin{aligned} S & = 1 + \frac{1}{4}\left(\frac{1}{2} + \frac{1}{3} \right) + \frac{1}{4^2}\left(\frac{1}{4} + \frac{1}{5} \right) + \frac{1}{4^3}\left(\frac{1}{6} + \frac{1}{7} \right) + ... \\ & = 2 \left(\frac{1}{2} + \frac{1}{3}\dot{}\left(\frac{1}{2}\right)^3 + \frac{1}{5}\dot{}\left(\frac{1}{2}\right)^5 +... \right) + \left(\frac{1}{2}\dot{}\left(\frac{1}{2}\right)^2 + \frac{1}{4}\dot{}\left(\frac{1}{2}\right)^4 + \frac{1}{6}\dot{}\left(\frac{1}{2}\right)^6 +... \right) \\ & = \ln \left(1+\frac{1}{2}\right) - \ln \left(1-\frac{1}{2}\right) - \frac{1}{2} \left(\ln \left(1+\frac{1}{2}\right) + \ln \left(1-\frac{1}{2}\right) \right) \quad \quad \small \color{#3D99F6}{\text{See Note.}} \\ & = \frac{1}{2} \ln \left(\frac{3}{2}\right) - \frac{3}{2} \ln \left(\frac{1}{2}\right) = \ln \left(\sqrt{\frac{3}{2}} \times 2 \sqrt{2} \right) = \ln \sqrt{12} \end{aligned}$

$\Rightarrow A = \boxed{12}$

$$

$\color{#3D99F6}{\text{Note:}}$

We know that for $0 \le x < 1$ , $\begin{cases} \ln(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + ... \\ \ln(1-x) = - x - \frac{1}{2}x^2 - \frac{1}{3}x^3 - \frac{1}{4}x^4 - ... \end{cases}$

$\begin{aligned} \Rightarrow \begin{cases} \ln(1+x) - \ln(1-x) & = 2 \left( x + \frac{1}{3}x^3 + \frac{1}{5}x^3 - \frac{1}{7}x^7 + ... \right) \\ - \ln(1+x) - \ln(1-x) & = 2 \left( \frac{1}{2}x^2 + \frac{1}{4}x^4 - \frac{1}{6}x^6 - \frac{1}{8}x^8 - ... \right) \end{cases} \end{aligned}$