Answer
$S = \dfrac{{47}}{{180}}$
Work Step by Step
We can write $\dfrac{1}{{n\left( {n + 3} \right)}}$ as
$\dfrac{1}{{n\left( {n + 3} \right)}} = \dfrac{1}{3}\left( {\dfrac{1}{n} - \dfrac{1}{{n + 3}}} \right)$
Thus, the partial sum becomes
${S_N} = \mathop \sum \limits_{n = 3}^N \dfrac{1}{{n\left( {n + 3} \right)}} = \dfrac{1}{3}\mathop \sum \limits_{n = 3}^N \left( {\dfrac{1}{n} - \dfrac{1}{{n + 3}}} \right)$
Since the summation index start from $3$, so we have the partial sums:
${S_3} = \dfrac{1}{3}\left( {\dfrac{1}{3} - \dfrac{1}{6}} \right)$
${S_4} = \dfrac{1}{3}\left[ {\left( {\dfrac{1}{3} - \dfrac{1}{6}} \right) + \left( {\dfrac{1}{4} - \dfrac{1}{7}} \right)} \right]$
${S_5} = \dfrac{1}{3}\left[ {\left( {\dfrac{1}{3} - \dfrac{1}{6}} \right) + \left( {\dfrac{1}{4} - \dfrac{1}{7}} \right) + \left( {\dfrac{1}{5} - \dfrac{1}{8}} \right)} \right]$
${S_6} = \dfrac{1}{3}\left[ {\left( {\dfrac{1}{3} - \dfrac{1}{6}} \right) + \left( {\dfrac{1}{4} - \dfrac{1}{7}} \right) + \left( {\dfrac{1}{5} - \dfrac{1}{8}} \right) + \left( {\dfrac{1}{6} - \dfrac{1}{9}} \right)} \right]$
${S_7} = \dfrac{1}{3}\left[ {\left( {\dfrac{1}{3} - \dfrac{1}{6}} \right) + \left( {\dfrac{1}{4} - \dfrac{1}{7}} \right) + \left( {\dfrac{1}{5} - \dfrac{1}{8}} \right) + \left( {\dfrac{1}{6} - \dfrac{1}{9}} \right) + \left( {\dfrac{1}{7} - \dfrac{1}{{10}}} \right)} \right]$
$\cdot\cdot\cdot$
In general, those fractions after and including $\dfrac{1}{6}$, that is, $\dfrac{1}{6},\dfrac{1}{7},\dfrac{1}{8},\cdot\cdot\cdot$ will be cancelled out. So, we obtain at the end:
$S = \dfrac{1}{3}\left( {\dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5}} \right)$
So, $S = \dfrac{{47}}{{180}}$.
