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

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Accepted Answer

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

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

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

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

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

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

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

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

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

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

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

$= 1$"

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

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

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