Seems interesting

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

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

Putting $x=\frac{1}{2}$ , then we have:

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

$\Rightarrow A = \dfrac{1+\frac{1}{2}}{1-\frac{1}{2}} = \boxed{3}$

look at the summation $\sum_{n=1}^\infty x^{2n-2} =\dfrac{1}{1-x^2}$ integrating both side we have $\sum_{n=1}^\infty \dfrac{x^{2n-1}}{2n-1}=\dfrac{\ln\left(\left|\dfrac{1+x}{1-x}\right|\right)}{2}$ divide both sides with x and put $x=\dfrac{1}{2}$ . $\sum_{n=1}^\infty \dfrac{1}{(2n-1)2^{2n-2}}=\dfrac{\ln\left(\left|\dfrac{1+\dfrac{1}{2}}{1-\dfrac{1}{2}}\right|\right)}{2*\dfrac{1}{2}}=\ln(\boxed{3})$

$Hey\ Here's\ my\ take$

$Let\ P = 1/2 + 1/{2^3}\cdot 3 + 1/{2^5}\cdot 5 +. . .$

$if\ we\ notice\ carefully\ we\ can\ see\ that\ it's\ equivalent\ to\ the\ following\ integral$

$I=\int\limits_0^\frac{1}{2}\ 1+x^2+x^4+. . .$

$applying\ the\ infinite\ geometric\ sum\ formula\ ;\ x\subset\ (0,1)$

$I=\int\limits_0^\frac{1}{2}\frac{1}{1-x^2} =ln\sqrt{3}$

$the\ integral\ being\ an\ easy\ one\ is\ left\ as\ an\ exercise\ for\ the\ reader$

$all\ we\ need\ is\ the\ value\ of\ 2I\ which\ is \ \boxed{ln3}$

It is very much intuitive as the series starts with 1+.... So it must be greater than 1;also notice the terms of the series decreases as we go on.Since 1+1/(3.2^2)=1.08333 ,the series should not increase so much after that(at least near to ln4=1.386..)so the only lnA remains greater than 1 is ln3.so it's ln3 & A=3.