Find the infinite sum tan − 1 ( 1/3 ) + tan − 1 ( 2/9 ) + … + tan − 1 ( 2 n − 1 / ( 1 + 2 2 n − 1 ) ) + …

The rth argument equals ( 2 r − 2 r − 1 ) / ( 1 + 2 r ⋅ 2 r − 1 ) , with r starting at 1. The partial sum is tan − 1 ( 2 n ) − tan − 1 1 ; the limit is π /2 − π /4 = π /4 .

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Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

MathematicsInverse Trigonometric FunctionsSummation of seriesMedium2 minQB

MathematicsMediumsingle choice

Passage / Comprehension

For positive adjacent values $z_{0}, z_{1}, \dots, z_{n}, tan^{- 1} ((z_{r} - z_{r - 1}) / (1 + z_{r} z_{r - 1})) = tan^{- 1} z_{r} - tan^{- 1} z_{r - 1}$ , so the sum telescopes.

Find the infinite sum $tan^{- 1} (1/3) + tan^{- 1} (2/9) + \dots + tan^{- 1} (2^{n - 1} / (1 + 2^{2 n - 1})) + \dots$

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Answer:

Solution:

The rth argument equals $(2^{r} - 2^{r - 1}) / (1 + 2^{r} \cdot 2^{r - 1})$ , with r starting at 1. The partial sum is $tan^{- 1} (2^{n}) - tan^{- 1} 1$ ; the limit is $π /2 - π /4 = π /4$ .

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

⏱ 2mℹ️ Source: QB

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