By using induction we will prove that $x_n \geq \frac{(n+1)(n+2)}{6}$. For $n=1$ the inequality holds. As $\sqrt{\frac{n(n+1)}{6}} \geq \frac{n+1}{3}$ (after squaring, the inequality is equivalent to $(n-2)(n+1) \geq 0$) we have: $x_{n}=x_{n-1}+\sqrt{x_{n-1}} \geq \frac{n(n+1)}{6} +\sqrt{\frac{n(n+1)}{6}} \geq \frac{n(n+1)}{6} + \frac{n+1}{3}=\frac{(n+1)(n+2)}{6}$. With which the induction is completed. Now we calculate $\sum_{n=1}^{2018}{\frac{1}{x_n}} \leq \sum_{n=1}^{2018}{\frac{6}{(n+1)(n+2)}}=\sum_{n=1}^{2018}(\frac{6}{n+1}-\frac{6}{n+2})=3 - \frac{6}{2020}<3$