lim n → ∞ tan ∑ r = 1 n tan - 1 1 1 + r + r 2 is equal to_______.

lim n → ∞ tan ∑ r = 1 n tan - 1 1 1 + r r + 1 = lim n → ∞ tan ∑ r = 1 n tan - 1 r + 1 - r 1 + r r + 1 = tan lim n → ∞ ∑ r = 1 n tan - 1 r + 1 - tan - 1 r = tan lim n → ∞ tan - 1 n + 1 - π 4 = tan π 4 = 1

Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

MathematicsInverse Trigonometric FunctionsSummation of seriesEasy2 minPYQ_2021

MathematicsEasynumerical

$\lim_{n \to \infty} \tan \{\sum_{r = 1}^{n} \tan^{- 1} (\frac{1}{1 + r + r^{2}})\}$ is equal to_______.

Answer:

1.00

Solution:

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

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

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

$= \tan (\lim_{n \to \infty} (\tan^{- 1} (n + 1) - \frac{π}{4}))$

$= \tan (\frac{π}{4}) = 1$

Stream:JEESubject:MathematicsTopic:Inverse Trigonometric FunctionsSubtopic:Summation of series

⏱ 2mℹ️ Source: PYQ_2021

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