(13) For any positive integer $n$, the sequence $\left\{a_{n}\right\}$ satisfies $\sum_{i=1}^{n} a_{i}=n^{3}$, then $\sum_{i=2}^{2009} \frac{1}{a_{i}-1}=$
$\sum_{i=2}^{2009} \frac{1}{a_{i}-1}=\sum_{i=2}^{2009} \frac{1}{3}\left(\frac{1}{i-1}-\frac{1}{i}\right)=\frac{1}{3}\left(1-\frac{1}{2009}\right)=\frac{2008}{6027}$.