Adding the inequalities
$$\begin{align*} a_n &\ge a_{n-1} + n + 1, \\ a_{n-1} &\ge a_{n-2} + n, \\ &\vdots, \\ a_2 &\ge a_1 + 3, \end{align*}$$
we obtain $a_n \ge a_1 + (3+4+\cdots+(n+1)) = \frac{n^2+3n+4}{2} > \frac{(n+1)(n+2)}{2}$. It follows
that
$$\sum_{n=1}^{2010} \frac{1}{a_n} < \sum_{n=1}^{\infty} \frac{2}{(n+1)(n+2)} = 2 \sum_{n=1}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+2} \right) = 2 \cdot \frac{1}{2} = 1 < \frac{3}{2}.$$