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.

Let $f(x)=\sum_{n=1}^{\infty} \frac{x^{2n+1}}{ n(2n+1)}$. Then we have $$f'(x)=\sum_{n=1}^{\infty} \frac{x^{2n}}{ n}=-\log(1-x^2).$$ Hence since f(0)=0, the sum is equal to
\begin{align}
s&=-\int_0^1\log(1-x^2)dx\\
&=-2\int_0^{\pi/2}\log(\cos x) \cos x dx\\
&=-2I
\end{align}
To solve this integral, $I$, note first that $\int_0^{\pi/2}\log(\sin x) \cos x dx=-1$. Thus
\begin{align}
I&=\int_0^{\pi/2}\log(\cos x) \cos x dx-\int_0^{\pi/2}\log(\sin x) \cos x dx+\int_0^{\pi/2}\log(\sin x) \cos x dx\\
&=\int_0^{\pi/2}\log(\cot x) \cos x dx-1\\
&=\int_0^{\pi/2}\Big(\log(\cot x) \cos x+\sec x -\sec x \Big)dx-1\\
&=\lim_{a\to \pi/2}\int_0^{a}\Big(\log(\cot x) \cos x-\sec x +\sec x \Big)dx-1\\
&=\lim_{a\to \pi/2}\Big(\log(\cot x) \sin x +\log [\cos \frac x2 + \sin \frac x2]-\log [\cos \frac x2 -\sin \frac x2] \Big)-1\\
&=\log2-1
\end{align} 
... I should still add more ... 
