$\lim _{n \rightarrow \infty} \tan \left\{\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{1+r+r^{2}}\right)\right\}$ is equal to _____.

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Inverse Trigonometric Functions

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

is equal to _____.

Answer: 1

Note that

\frac{1}{1 + r + r ^{2}} = \frac{1}{r ( r + 1 ) + 1}

.

Using the identity

tan^{- 1} (a) - tan^{- 1} (b) = tan^{- 1} (\frac{a - b}{1 + ab})

, we can write:

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

So the sum telescopes:

r = 1 \sum n [tan^{- 1} (r + 1) - tan^{- 1} (r)] = tan^{- 1} (n + 1) - tan^{- 1} (1)

n \to \infty

tan^{- 1} (n + 1) \to \frac{π}{2}

.

Thus:

n \to \infty lim tan [tan^{- 1} (n + 1) - tan^{- 1} (1)] = tan (\frac{π}{2} - \frac{π}{4}) = tan (\frac{π}{4}) = 1

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