College Algebra, Chapter 9, 9.5, Section 9.5, Problem 12

Prove that the formula $\displaystyle \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + ... + \frac{1}{n (n + 1)} = \frac{n}{(n + 1)}$ is true for all natural numbers $n$.

By using mathematical induction,

Let $P(n)$ denote the statement $\displaystyle \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + ... + \frac{1}{n (n + 1)} = \frac{n}{(n + 1)}$.

Then, we need to show that $P(1)$ is true. So,


$\begin{equation} \begin{aligned} \frac{1}{1 \cdot 2} =& \frac{1}{( 1 + 1)} \\ \\ \frac{1}{2} =& \frac{1}{2} \end{aligned} \end{equation}$


Thus, we prove the first principle of the mathematical induction. More over, assuming that $P(k)$ is true, then

$\displaystyle \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + ... + \frac{1}{k (k + 1)} = \frac{k}{(k + 1)}$

Now, by showing $P(k + 1)$, we have


$\begin{equation} \begin{aligned} \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + ... + \frac{1}{k (k + 1)} + \frac{1}{(k + 1) [(k + 1) + 1]} =& \frac{k + 1}{[(k + 1) + 1]} \\ \\ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + ... + \frac{1}{k (k + 1)} + \frac{1}{(k + 1)(k + 2)} =& \frac{k + 1}{k + 2} \end{aligned} \end{equation}$


We start with the left side and use the induction hypothesis to obtain the right side of the equation:


$\begin{equation} \begin{aligned} =& \left[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + ... + \frac{1}{k (k + 1)} \right] + \left[ \frac{1}{(k + 1)(k + 2)} \right] && \text{Group the first $k$ terms} \\ \\ =& \frac{k}{k + 1} + \frac{1}{(k + 1)(k + 2)} && \text{Induction hypothesis} \\ \\ =& \frac{k (k + 2) + 1}{(k + 1)(k + 2)} && \text{Get the LCD} \\ \\ =& \frac{k^2 + 2k + 1}{(k + 1)(k + 2)} && \text{Expand the numerator} \\ \\ =& \frac{( k + 1)^2}{(k + 1)(k + 2)} && \text{Factor} \\ \\ =& \frac{k + 1}{k + 2} && \text{Simplify} \end{aligned} \end{equation}$


Therefore, $P(k+1)$ follows from $P(k)$, and this completes the induction step.