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.

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 $$\sum_{n=1}^\infty \frac{1}{n(2n+1)} =\lim_{N\to\infty}\sum_{n=1}^N \frac{1}{n(2n+1)} = 2 \int_0^1 \frac{z}{1+z} dz = 2 (1 - \log 2)$$

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.

$$\frac{1}{n(2n+1)}= \frac 1n-\frac{2}{2n+1}=2\left(\frac{1}{2n}-\frac{1}{2n+1}\right)$$

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 $${\displaystyle \ln(1+x)=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}x^{n}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots ,}$$