## College Algebra 7th Edition

$\sum_{k=1}^{999}\frac{1}{k(k+1)}$
We write the sum in sigma notation: (We notice that the denominator takes the form of $n(n+1)$ with $n$ being consecutive integers.) $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots+\frac{1}{999\cdot 1000}=\sum_{k=1}^{999}\frac{1}{k(k+1)}$