The value of $\lim _{n \rightarrow \infty} 6 \tan \left\{\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{r^{2}+3 r+3}\right)\right\}$ is equal to

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Inverse Trigonometric Functions

The value of

lim_{n \to \infty} 6 tan {\sum_{r = 1}^{n} tan^{- 1} (\frac{1}{r ^{2} + 3 r + 3})}

is equal to

Answer: (C)

Write

T_{r} = tan^{- 1} (\frac{1}{r ^{2} + 3 r + 3}) = tan^{- 1} (\frac{( r + 2 ) - ( r + 1 )}{1 + ( r + 2 ) ( r + 1 )}) = tan^{- 1} (r + 2) - tan^{- 1} (r + 1)

. This is a telescoping series:

S_{n} = \sum_{r = 1}^{n} T_{r} = tan^{- 1} (n + 2) - tan^{- 1} (2) = tan^{- 1} (\frac{n}{2 n + 5})

. Therefore,

lim_{n \to \infty} 6 tan (S_{n}) = lim_{n \to \infty} \frac{6 n}{2 n + 5} = \frac{6}{2} = 3

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