Answer
$\begin{array}{*{20}{c}}
N&{{S_N}}\\
1&{\dfrac{2}{3}}\\
2&{\dfrac{{11}}{{12}}}\\
3&{\dfrac{{21}}{{20}}}\\
4&{\dfrac{{17}}{{15}}}
\end{array}$
$S = \mathop {\lim }\limits_{N \to \infty } {S_N} = \dfrac{3}{2}$
Work Step by Step
Given $S = \mathop \sum \limits_{n = 1}^\infty \left( {\dfrac{1}{n} - \dfrac{1}{{n + 2}}} \right)$, we compute ${S_N}$ for $N = 1,2,3,4$:
${S_1} = 1 - \dfrac{1}{3} = \dfrac{2}{3}$
${S_2} = \left( {1 - \dfrac{1}{3}} \right) + \left( {\dfrac{1}{2} - \dfrac{1}{4}} \right)$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ ${S_2} = 1 + \dfrac{1}{2} - \dfrac{1}{3} - \dfrac{1}{4} = \dfrac{{11}}{{12}}$
${S_3} = \left( {1 - \dfrac{1}{3}} \right) + \left( {\dfrac{1}{2} - \dfrac{1}{4}} \right) + \left( {\dfrac{1}{3} - \dfrac{1}{5}} \right)$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ ${S_3} = 1 + \dfrac{1}{2} + \left( { - \dfrac{1}{3} + \dfrac{1}{3}} \right) - \dfrac{1}{4} - \dfrac{1}{5} = \dfrac{{21}}{{20}}$
${S_4} = \left( {1 - \dfrac{1}{3}} \right) + \left( {\dfrac{1}{2} - \dfrac{1}{4}} \right) + \left( {\dfrac{1}{3} - \dfrac{1}{5}} \right) + \left( {\dfrac{1}{4} - \dfrac{1}{6}} \right)$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ ${S_4} = 1 + \dfrac{1}{2} + \left( { - \dfrac{1}{3} + \dfrac{1}{3}} \right) + \left( { - \dfrac{1}{4} + \dfrac{1}{4}} \right) - \dfrac{1}{5} - \dfrac{1}{6} = \dfrac{{17}}{{15}}$
Here, we notice a pattern:
${S_N} = 1 + \dfrac{1}{2} + \left( { - \dfrac{1}{3} + \dfrac{1}{3}} \right) + \left( { - \dfrac{1}{4} + \dfrac{1}{4}} \right) + \cdot\cdot\cdot - \dfrac{1}{{N + 1}} - \dfrac{1}{{N + 2}}$
Since the pairs in the brackets are zero, we obtain
${S_N} = \dfrac{3}{2} - \dfrac{1}{{N + 1}} - \dfrac{1}{{N + 2}}$
By definition: $S = \mathop {\lim }\limits_{N \to \infty } {S_N}$. Thus,
$S = \mathop {\lim }\limits_{N \to \infty } {S_N} = \mathop {\lim }\limits_{N \to \infty } \left( {\dfrac{3}{2} - \dfrac{1}{{N + 1}} - \dfrac{1}{{N + 2}}} \right) = \dfrac{3}{2}$
