## Precalculus (6th Edition) Blitzer

The simplified solution is $\frac{1}{x}-\frac{1}{x+1}$ and its sum is $\frac{99}{100}$.
Let us consider the rational expression. \begin{align} & \frac{1}{x\left( x+1 \right)}=\frac{A}{x}+\frac{B}{x+1} \\ & =\frac{A\left( x+1 \right)+Bx}{x\left( x+1 \right)} \end{align} $1=A\left( x+1 \right)+Bx$ (I) Substitute the value of $x=-1$ in the equation (I), \begin{align} & 1=B\left( -1 \right) \\ & B=-1 \end{align} Again, putting the value of $x=0$ in the equation (I), \begin{align} & 1=A\left( 1 \right) \\ & A=1 \end{align} Therefore, $\frac{1}{x\left( x+1 \right)}=\frac{1}{x}-\frac{1}{x+1}$ is the simplified form of the result By using the above result to find the sum of the series, \begin{align} & \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{99\cdot 100} \\ & =\left( \frac{1}{1}-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+\cdots +\left( \frac{1}{99}-\frac{1}{100} \right) \\ & =\frac{1}{1}-\frac{1}{100} \\ & =\frac{99}{100} \end{align} Thus, the sum is $\frac{99}{100}$.