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Evaluate: _{ r=1 }^{ }{ ^{ -1 }{ ( { 2 }{ 1+(2r+1)(2r-1)

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Evaluate: $\sum_{r = 1}^{\infty} tan^{- 1} (\frac{2}{1 + ( 2 r + 1 ) ( 2 r - 1 )})$

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Evaluate:

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

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Text solutionVerified

\sum_{r = 1}^{\infty} t a n^{- 1} (\frac{2}{1 + ( 2 r + 1 ) ( 2 r - 1 )})

Divide numerator and denomenator by

(2 r + 1) (2 r - 1)

\sum_{r = 1}^{\infty} t a n^{- 1} ⎝ ⎛ \frac{\frac{2}{( 2 r + 1 ) ( 2 r - 1 )}}{\frac{1}{( 2 r + 1 ) ( 2 r - 1 )} + 1} ⎠ ⎞

\sum_{r = 1}^{\infty} t a n^{- 1} ⎝ ⎛ \frac{\frac{1}{2 r - 1} - \frac{1}{2 r + 1}}{\frac{1}{( 2 r + 1 ) ( 2 r - 1 )} + 1} ⎠ ⎞

t a n^{- 1} \frac{a - b}{1 + ab} = t a n^{- 1} a - t a n^{- 1} b

∴

\sum_{r = 1}^{\infty} t a n^{- 1} ⎝ ⎛ \frac{\frac{1}{2 r - 1} - \frac{1}{2 r + 1}}{\frac{1}{( 2 r + 1 ) ( 2 r - 1 )} + 1} ⎠ ⎞ = \sum_{r = 1}^{\infty} (t a n^{- 1} \frac{1}{2 r - 1} - t a n^{- 1} \frac{1}{2 r + 1})

Open the summation we get,

(t a n^{- 1} 1 - t a n^{- 1} \frac{1}{3}) + (t a n^{- 1} \frac{1}{3} - t a n^{- 1} \frac{1}{5}) + (t a n^{- 1} \frac{1}{5} - t a n^{- 1} \frac{1}{7}) ...........\infty

At infinity

t a n^{- 1} \frac{1}{\infty} = 0

And other terms all cancel out each other, only

t a n^{- 1} 1

remains.

∴

Ans is

t a n^{- 1} 1 = \frac{π}{4}

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Question Text	Evaluate: $\sum_{r = 1}^{\infty} tan^{- 1} (\frac{2}{1 + ( 2 r + 1 ) ( 2 r - 1 )})$
Answer Type	Text solution:1
Upvotes	150