Please help, Thank YOU if you try

$\frac{1}{\frac{1(1+1)}{2}}+\frac{1}{\frac{2(2+1)}{2}}+\dots + \frac{1}{\frac{2018(2018+1)}{2}}$

$\frac{2}{1(2)}+\frac{2}{2(3)}+ \dots+\frac{2}{2018(2019)}$

$2(\frac{1}{1(2)}+ \frac{1}{2(3)} + \dots + \frac{1}{2018(2019)})$
$\text{Apply } \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

$2(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3} +\dots -\frac{1}{2019})$

$2(1-\frac{1}{2019})$

$\frac{4036}{2019}$