## Precalculus (6th Edition) Blitzer

We use mathematical induction as follows: Statement ${{S}_{1}}$ is; $\frac{1}{1\cdot 2}=\frac{1}{2}$ And simplifying on the right, we obtain $\frac{n}{n+1}=\frac{1}{2}$. So, this true statement shows that ${{S}_{1}}$ is true. Also, assume ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression, $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{k\left( k+1 \right)}=\frac{k}{k+1}$ By adding $\frac{1}{\left( k+1 \right)\left( k+2 \right)}$ on both sides as: \begin{align} & \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\ldots +\frac{1}{\left( k+1 \right)\left( k+2 \right)}=\frac{k}{k+1}+\frac{1}{\left( k+1 \right)\left( k+2 \right)} \\ & \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\ldots +\frac{1}{\left( k+1 \right)\left( k+2 \right)}=\frac{k\left( k+2 \right)}{k+1\left( k+2 \right)}+\frac{1}{\left( k+1 \right)\left( k+2 \right)} \\ & \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\ldots +\frac{1}{\left( k+1 \right)\left( k+2 \right)}=\frac{k\left( k+2 \right)+1}{k+1\left( k+2 \right)} \end{align} Therefore, \begin{align} & \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\ldots +\frac{1}{\left( k+1 \right)\left( k+2 \right)}=\frac{{{k}^{2}}+2k+1}{k+1\left( k+2 \right)} \\ & \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\ldots +\frac{1}{\left( k+1 \right)\left( k+2 \right)}=\frac{\left( k+1 \right)}{\left( k+2 \right)} \end{align} Thus, ${{S}_{k+1}}$ is true. The result $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\ldots +\frac{1}{n\left( n+1 \right)}=\frac{n}{n+1}$ holds true.