Think it out

$ln(\frac{1+x}{1-x})=2x+\frac{2}{3}x^{3}+\frac{2}{5}x^{5}+\frac{2}{7}x^{7}........$

Replace $x=\frac{1}{2}$

$Answer=ln3$

The series can be written as

$\displaystyle \sum_{n=1}^\infty \frac{1}{2^{2n-2}\cdot(2n-1)}\Rightarrow \displaystyle 2\sum_{n=1}^\infty \frac{1}{2^{2n-1}\cdot(2n-1)}$

The latter expression resembles that of the Maclaurin series representation of $\tanh^{-1} (x)$ .

$\tanh^{-1} (x)=\displaystyle \sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)} \\ \tanh^{-1} (\frac{x}{2})=\displaystyle \sum_{n=1}^\infty \frac{x^{2n-1}}{2^{2n-1}\cdot(2n-1)}$

Hence, with $x=1$

$\displaystyle \sum_{n=1}^\infty \frac{1}{2^{2n-2}\cdot(2n-1)}=2\tanh^{-1} (0.5)\approx \boxed{1.0986}$

Note: By applying the properties of hyperbolic functions, it can be shown that $\tanh^{-1}(0.5)=\frac{\ln{3}}{2}$