Log in a sum

$\displaystyle \sum_{n=0}^{\infty} \dfrac{2^{-2n}}{2n+1} \\ \displaystyle =\sum_{n=0}^{\infty} \int_{0}^{1} x^{2n} 2^{-2n} dx \\ \displaystyle =\int_{0}^{1} \sum_{n=0}^{\infty} (\dfrac{x}{2})^{2n} dx \displaystyle =\int_{0}^{1} \dfrac{1}{1-(x/2)^2} dx=\ln {3}$

From Maclaurin series , we have:

$\begin{aligned} \ln(1+x) & = x - \frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4 + \cdots & -1 < x \le 1 \\ \ln(1-x) & = -x - \frac {x^2}2 - \frac {x^3}3 - \frac {x^4}4 - \cdots \\ \ln(1+x) - \ln(1-x) & = 2x + \frac {2x^3}3 + \frac {2x^5}5 + \frac {2x^7}7 + \cdots \\ \implies \ln \left( \frac {1+x}{1-x}\right) & = \sum_{n=0}^\infty \frac {2x^{2n+1}}{2n+1} & \small \color{#3D99F6} \text{Putting }x= \frac 12 \\ \ln \left( \frac {1+\frac 12}{1-\frac 12}\right) & = \sum_{n=0}^\infty \frac {2\cdot2^{-(2n+1)}}{2n+1} \\ & = \sum_{n=0}^\infty \frac {2^{-2n}}{2n+1} = a \\ \implies a & = \ln \left( \frac {1+\frac 12}{1-\frac 12}\right) = \ln 3 \end{aligned}$

$\implies e^a = e^{\ln 3} = \boxed{3}$