What is the value of $\frac {1}{2\cdot3} + \frac {1}{4\cdot 5} +\frac {1}{6\cdot 7} + \cdots $? What is value of $\dfrac {1}{2\cdot 3} + \dfrac {1}{4\cdot 5} +\dfrac {1}{6\cdot7} + \cdots  =?$
1) $\log \left( \dfrac 2e \right)$
2) $\log \left( \dfrac e2 \right)$
3) $\log \left( 2e \right)$
4) $e-1$
I know it converges but I am not getting how to solve 

An alternative solution, just for fun:
$$\begin{eqnarray*}\sum_{n\geq 1}\frac{1}{2n(2n+1)} = \int_{0}^{1}\sum_{n\geq 1}\frac{x^{2n}}{2n}\,dx &=& -\frac{1}{2}\int_{0}^{1}\log(1-x^2)\,dx \\(IBP)\quad&=&\int_{0}^{1}x\cdot\frac{1-x}{1-x^2}\,dx\\&=&\int_{0}^{1}\frac{x}{1+x}\,dx\\&=&1-\log 2=\color{red}{\log\left(\frac{e}{2}\right)}.\end{eqnarray*}$$
