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.
 A: First, we can rewrite the partial sum as an integral
$$\sum_{n=1}^N \frac{1}{n(2n+1)} = 2\sum_{n=1}^N \left(\frac{1}{2n} - \frac{1}{2n+1}\right)
= 2\sum_{n=1}^N \int_0^1 (z^{2n-1} - z^{2n}) dz\\
= 2 \int_0^1 z(1-z)\left(\sum_{n=0}^{N-1} z^{2n}\right) dz
= 2 \int_0^1 \frac{z}{1+z}( 1 - z^{2N} ) dz
$$
Notice the $N$ dependence piece on RHS can be bounded from above
$$\left| 2 \int_0^1 \frac{z}{1+z} z^{2N} dz \right|
< 2 \int_0^1 z^{2N} dz = \frac{2}{2N+1} \to 0 
\quad\text{ as }\quad N \to \infty
$$
We have
$$\sum_{n=1}^\infty \frac{1}{n(2n+1)} 
=\lim_{N\to\infty}\sum_{n=1}^N \frac{1}{n(2n+1)} 
= 2 \int_0^1 \frac{z}{1+z} dz = 2 (1 - \log 2)$$
A: The sum can be found using our favourite alternative method of converting the sum into a double integral.
Noting that
$$\int_0^1 x^{n - 1} \, dx = \frac{1}{n} \quad \text{and} \quad \int_0^1 y^{2n} \, dy = \frac{1}{2n + 1},$$
the sum can be rewritten as
\begin{align*}
\sum_{n = 1}^\infty \frac{1}{n (2n + 1)} &= \sum_{n = 1}^\infty \int_0^1 \int_0^1 x^{n - 1} y^{2n} \, dx dy\\
&= \int_0^1 \int_0^1 \sum_{n = 1}^\infty x^{n - 1} y^{2n} \, dx dy \tag1 \\
&= \int_0^1 \int_0^1 \frac{1}{x} \sum_{n = 1}^\infty (xy^2)^n \, dx dy\\
&= \int_0^1 \int_0^1 \frac{1}{x} \cdot \frac{xy^2}{1 - xy^2} \, dx dy \tag2\\
&= \int_0^1 \int_0^1 \frac{y^2}{1 - xy^2} \, dx dy\\
&= -\int_0^1 \ln (1 - xy^2) \Big{|}_0^1 \, dy\\
&= - \int_0^1 \ln (1 - y^2) \, dy\\
&= 2 \int_0^1 \frac{y(1 - y)}{1 - y^2} \, dy \tag3\\
&= 2 \int_0^1 \frac{y}{1 + y} \, dy\\
&= 2 \int_0^1 \frac{(1 + y) - 1}{1 + y} \, dy\\
&= 2 \int_0^1 \left (1 - \frac{1}{1 + y} \right ) \, dy\\
&= 2 \left [y - \ln (1 + y) \right ]_0^1\\
&= 2 (1 - \ln 2).
\end{align*}
Explanation
(1) Changing the summation with the double integration.
(2) Summing the series which is geometric.
(3) Integrating by parts.
A: $$\frac{1}{n(2n+1)}= \frac 1n-\frac{2}{2n+1}=2\left(\frac{1}{2n}-\frac{1}{2n+1}\right)$$
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 $${\displaystyle \ln(1+x)=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}x^{n}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots ,}$$
A: 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 ... 
A: $$\sum_{n=1}^\infty\frac1{n(2n+1)}=\sum_{n=1}^\infty\frac1n\int_0^1 x^{2n}dx=\int_0^1\sum_{n=1}^\infty\frac{x^{2n}}{n}dx=-\int_0^1\ln(1-x^2)dx$$
$$=-\underbrace{\int_0^1\ln(1-x)dx}_{1-x=y}-\underbrace{\int_0^1\ln(1+x)dx}_{1+x=y}=-\int_0^2\ln(y)dy$$
$$=y-y\ln(y)|_0^2=2-2\ln(2)$$
A: Your series can be written as 
$$
    \frac{H_{x}}{2x} = \sum_{k=1}^\infty \frac{1}{k\cdot2(k+x)}\, ,
$$
with $x=\frac12$. $H_{\frac12}$ is given here: Harmonic numbers.
$$
    H_{\frac{1}{2}} = 2 -2\ln{2},
$$
which is your result...
