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 

We use $$\frac{1}{n(n+1)}=\frac{n+1-n}{n(n+1)}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$
Note that your series, $$\dfrac {1}{2 \times 3} + \dfrac {1}{4 \times 5} +\dfrac {1}{6 \times 7} + \cdots=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\dots$$
Since $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots=\ln 2$$
This result is known, and is discussed as the Alternating Harmonic Series on Wikipedia. So since $$\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\dots=1-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots \right)$$ the answer is $1-\ln 2$. So the answer to your question is $$1-\ln 2=\ln e-\ln 2=\ln \frac{e}{2}$$So the answer is $2)$
