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 


Herein we present a methodology that relies on only straightforward arithmetic and knowledge of Riemann sums.  To that end, we now proceed.


The general term in the series is $a_n=\frac{1}{2n(2n+1)}=\frac{1}{2n}-\frac{1}{2n+1}$.
Then, we have
$$\begin{align}
\sum_{n=1}^N a_n&=\sum_{n=1}^{N}\left(\frac{1}{2n}-\frac{1}{2n+1}\right)\\\\
&=-\sum_{n=1}^{N}\left(\frac{1}{2n}+\frac{1}{2n+1}\right)+\sum_{n=1}^N\frac1n\\\\
&=1-\left(\sum_{n=1}^{2N+1}\frac1n-\sum_{n=1}^{N}\frac1n\right)\\\\
&=1-\sum_{n=N+1}^{2N+1}\frac1n\\\\
&=1-\sum_{n=1}^{N+1}\frac1{n+N}\\\\
&=1-\frac1N\sum_{n=1}^{N+1}\frac{1}{1+n/N}\tag 1
\end{align}$$
Recognizing the sum on the right-hand side of $(1)$ as the Riemann sum for $\int_0^1 \frac{1}{1+x}\,dx=\log(2)$, we have 

$$\lim_{N\to \infty}\sum_{n=1}^Na_n=1-\log(2)=\log(e/2)$$

