9. (8-9 grade) Compare the numbers $\frac{1}{2!}+\frac{2}{3!}+\ldots+\frac{2012}{2013!}$ and 1. ( $n$ ! denotes the product $1 \cdot 2 \cdot \ldots \cdot n$ )
Answer: The first number is smaller.

Solution. Note that $\frac{n}{(n+1)!}=\frac{(n+1)}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$. When adding, the inner terms will cancel out, leaving $1-\frac{1}{2013!}$.