Compute the following: $\displaystyle \sum_{n = 1}^3 \left( \dfrac{n + 1}{n} \right) - \sum_{n = 1}^3 \left( \dfrac{n}{n + 1} \right)$.

$\begin{align}
\displaystyle \sum_{n = 1}^3 \left( \dfrac{n + 1}{n} \right) - \sum_{n = 1}^3 \left( \dfrac{n}{n + 1} \right) & = \left( \dfrac{2}{1} + \dfrac{3}{2} + \dfrac{4}{3} \right) - \left( \dfrac{1}{2} + \dfrac{2}{3} + \dfrac{3}{4} \right) \\
& = \dfrac{29}{6} - \dfrac{23}{12} \\
\\
& = \dfrac{35}{12}
\end{align}$