7.3. Find the sum $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{2020 \cdot 2021}$.
Answer: $\frac{2020}{2021}$.

Solution. Notice that $\frac{1}{n \cdot(n+1)}=\frac{1}{n}-\frac{1}{n+1}$. Then the sum equals $1-\frac{1}{2021}=\frac{2020}{2021}$.