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**Prove It** Notice that $\displaystyle \displaystyle \begin{align*} \frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1} \end{align*}$, so your sum

$\displaystyle \displaystyle \begin{align*} \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n + 1)} &= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4} \right) + \dots + \left(\frac{1}{n} - \frac{1}{n + 1}\right) \\ &= 1 - \frac{1}{n + 1} \\ &= \frac{n + 1}{n + 1} - \frac{1}{n + 1} \\ &= \frac{n}{n+1} \end{align*}$