The value of [ _n 6 _r = 1^n ^ - 1( 1 r^2 + 3r + 3 ) ] is equal to :

Question

The value of

$$\mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {{r^2} + 3r + 3}}} \right)} } \right\}$$ is equal to :

Accepted Answer

$$\mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {{r^2} + 3r + 3}}} \right)} } \right\}$$

$$ = \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{{(r + 2) - (r + 1)} \over {1 + (r + 2)(r + 1)}}} \right)} } \right\}$$

$$ = \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {({{\tan }^{ - 1}}(r + 2) - {{\tan }^{ - 1}}(r + 1))} } \right\}$$

$$ = \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {{{\tan }^{ - 1}}(n + 2) - {{\tan }^{ - 1}}2} \right\}$$

$$ = 6\tan \left\{ {{\pi  \over 2} - {{\cot }^{ - 1}}\left( {{1 \over 2}} \right)} \right\}$$

$$ = 6\tan \left( {{{\tan }^{ - 1}}\left( {{1 \over 2}} \right)} \right)$$

$= 3$

Solutions · 1