Sum this 2015

$\begin{aligned} \sum_{n=1}^{2015} \frac{16n-48}{(n+6)(n+3)(n)} & = \sum_{n=1}^{2015} \left( \frac{16}{(n+3)(n+6)} - \frac{48}{n(n+3)(n+6)} \right) \\ & = \sum_{n=1}^{2015} \left( \frac{16}{3} \left[ \frac{1}{n+3} - \frac{1}{n+6} \right] - \frac{48}{18} \left[ \frac{1}{n} - \frac{2}{n+3} + \frac{1}{n+6} \right] \right) \\ & = \frac{8}{3} \sum_{n=1}^{2015} \left( - \frac{1}{n} + \frac{4}{n+3} - \frac{3}{n+6} \right) \\ & = \frac{8}{3} \left(- \sum_{n=1}^{2015} \frac{1}{n} + \sum_{n=1}^{2015} \frac{1}{n+3} + 3 \sum_{n=1}^{2015} \frac{1}{n+3}- 3 \sum_{n=1}^{2015} \frac{1}{n+6} \right) \\ & = \frac{8}{3} \left(- \sum_{n=1}^{2015} \frac{1}{n} + \sum_{n=4}^{2018} \frac{1}{n} + 3 \left[\sum_{n=4}^{2018} \frac{1}{n} - \sum_{n=7}^{2021} \frac{1}{n} \right] \right) \\ & = \frac{8}{3} \left( - \frac{1}{1} - \frac{1}{2} - \frac{1}{3} + \frac{1}{2016} + \frac{1}{2017} + \frac{1}{2018} + 3\left[\frac{1}{4} + \frac{1}{5} + \frac{1}{6} - \frac{1}{2019} - \frac{1}{2020} - \frac{1}{2021} \right] \right) \\ & = \boxed{0.037} \end{aligned}$