$${\tan ^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)$$

$$ = {\tan ^{ - 1}}\left( {{{r + 1 - r} \over {1 + r(r + 1)}}} \right)$$

$$ = {\tan ^{ - 1}}(r + 1) - {\tan ^{ - 1}}r$$

$$ \therefore $$ $$\sum\limits_{r = 1}^n {\left( {{{\tan }^{ - 1}}(r + 1) - {{\tan }^{ - 1}}(r)} \right)} $$

$$ = {\tan ^{ - 1}}(2) - {\tan ^{ - 1}}(1) + ta{n^{ - 1}}(3) - {\tan ^1}(2) + ta{n^{ - 1}}(n + 1) - {\tan ^{ - 1}}(n)$$

$$ = {\tan ^{ - 1}}(n + 1) - {\tan ^{ - 1}}(1)$$

$$ = {\tan ^{ - 1}}\left( {{{n + 1 - 1} \over {1 + (n + 1)1}}} \right)$$

$$ = {\tan ^{ - 1}}\left( {{n \over {n + 2}}} \right)$$

$$\mathop {\lim }\limits_{n \to \infty } \tan \left( {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)} } \right)$$

$$ = \mathop {\lim }\limits_{x \to \infty } \tan \left( {{{\tan }^{ - 1}}\left( {{n \over {n + 2}}} \right)} \right)$$

$$ = \mathop {\lim }\limits_{x \to \infty } {n \over {n + 2}}$$

$$ = 1$$