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