7. Given $a_{n}=\log _{n}(n+1)$, let $\sum_{n=2}^{1023} \frac{1}{\log _{a_{n}} 100}=\frac{q}{p}$, where $p, q$ are positive integers, and $(p, q)=1, p+q=$ $\qquad$
7. 3 .
$a_{n}=\log _{n}(n+1)=\frac{\lg (n+1)}{\lg n}$, then $a_{2} a_{3} \cdots a_{n}=\frac{\lg 3}{\lg 2} \cdot \frac{\lg 4}{\lg 3} \cdots \frac{\lg 1024}{\lg 1023}=\frac{\lg 1024}{\lg 2}=10$, hence $\sum_{n=2}^{1023} \frac{1}{\log _{a_{n}} 100}=\sum_{n=2}^{1023} \log _{100} a_{n}=\log _{100}\left(a_{2} a_{3} \cdots a_{1023}\right)=\log _{100} 10=\frac{1}{2}=\frac{q}{p}$. Also, $(p, q)=1$, then $p=2, q=1$