<s>[INST] 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.
 [/INST] 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.
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