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{ }_{n } _{r = 1}^n {{{ }^{ - 1}}( {{{2r}}{{1 - {r^2} + {r^

Question

$n \to \infty lim r = 1 \sum n tan^{- 1} (\frac{2 r}{1 - r ^{2} + r ^{4}})$ is equal to

A

\frac{π}{4}

B

\frac{π}{2}

C

\frac{3 π}{4}

D

none of these

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Theodore Discussed

n \to \infty lim r = 1 \sum n tan^{- 1} (\frac{2 r}{1 - r ^{2} + r ^{4}})

is equal to

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Text solutionVerified

lim n \to \infty \sum_{r = 1}^{n} tan^{- 1} (\frac{2 r}{1 - r ^{2} + r ^{4}})

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

= lim n \to \infty \sum_{r = 1}^{n} tan^{- 1} (\frac{r ( r + 1 ) - r ( r - 1 )}{1 + [ r ( r - 1 )] [ r ( r + 1 )]})

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

= lim n \to \infty [t a n^{- 1} (1 (1 + 1) - tan^{- 1} (1 (1 - 1)) + t a n^{- 1} (2 (2 + 1)) - tan^{- 1} (2 (2 - 1)) + ....... + tan^{- 1} (n (n + 1)) - tan^{- 1} (n (n - 1))]

= lim n \to \infty [t a n^{- 1} (2) - tan^{- 1} (0) + t a n^{- 1} (6)) - tan^{- 1} (2) + ......... + tan^{- 1} (n (n + 1)) - tan^{- 1} (n (n - 1))]

= lim n \to \infty [- tan^{- 1} (0) + tan^{- 1} (n (n + 1))]

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

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

= tan^{- 1} (\infty)

= \frac{π}{2}

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Question Text	$n \to \infty lim r = 1 \sum n tan^{- 1} (\frac{2 r}{1 - r ^{2} + r ^{4}})$ is equal to
Answer Type	Text solution:1
Upvotes	150