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.