How to find sum of the infinite series $\sum_{n=1}^{\infty} \frac{1}{ n(2n+1)}$ $$\frac{1}{1 \times3} + \frac{1}{2\times5}+\frac{1}{3\times7} + \frac{1}{4\times9}+\cdots $$
How to find sum of this series?
I tried this: its $n$th term will be = $\frac{1}{n}-\frac{2}{2n+1}$; after that  I am not able to solve this.

$$\frac{1}{n(2n+1)}= \frac 1n-\frac{2}{2n+1}=2\left(\frac{1}{2n}-\frac{1}{2n+1}\right)$$
Hence 
$$\sum_{n=1}^{\infty} \frac{1}{ n(2n+1)} =2 \sum_{n=1}^{\infty} \left(\frac{1}{2n}-\frac{1}{2n+1}\right)= 2\left(1-\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\right)=\color{red}{ 2(1-\ln 2)} $$
since $${\displaystyle \ln(1+x)=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}x^{n}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots ,}$$
