Find the sum of the infinite series

Question

Find the sum of the infinite series: $$ \begin{align*}\frac{1}{2.3.4}+\frac{1}{4.5.6} +\frac{1}{6.7.8}+\frac{1}{8.9.10} + \ldots &= \sum_{n=1}^{\infty}\frac{1}{2n(2n+1)(2n+2))} \\ &= \frac{1}{4}\sum_{1}^{\infty}\frac{1}{n(n+1)(2n+1)} \\ &=\sum_{1}^{\infty}\frac{1}{4n}-\sum_{1}^{\infty}\frac{1}{2n+1}+\sum_{1}^{\infty}\frac{1}{2(2n+2)} \\ &= \frac14+\frac14+\frac16+\ldots 1-\frac12-\frac14-\frac16-\ldots -1+\frac12-\frac13+\frac14 \ldots \\ &=1+\frac14-\frac12-\ln2 \\ &=\frac{1}{4}[3-4\ln2] \end{align*} $$

Answer 1

\begin{equation} \sum_{n=1}^{\infty}\frac{1}{2n(2n+1)(2n+2))}=\frac{1}{4}\sum_{1}^{\infty}\frac{1}{n(n+1)(2n+1)}=\frac{1}{4}[3-4\ln2] \end{equation}