Base case: \(n=1\)
\[\frac{1}{1\cdot2}=\frac{1}{2}\]
\[1-\frac{1}{1+1}=1-\frac{1}{2}=\frac{1}{2}\]
So \(\sum_{k=1}^{n} \frac{1}{k(k+1)}=1-\frac{1}{n+1}\) for \(n=1\)
Induction step:
\[\begin{align}
\sum_{k=1}^{n+1} \frac{1}{k(k+1)} &= \sum_{k=1}^{n} \frac{1}{k(k+1)}+\frac{1}{(n+1)(n+2)}\\
&= 1-\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}\\
&= 1-\frac{n+2-1}{(n+1)(n+2)}\\
&= 1-\frac{1}{n+2}
\end{align}\]
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