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 
 A: 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 $2)$
A: 
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 

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

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