The sum of the infinite series cot – 1 ( 7 4 ) + cot – 1 ( 19 4 ) + cot – 1 ( 39 4 ) + cot – 1 ( 67 4 ) + . . . is: [2025]

Question

The sum of the infinite series $\cot^{- 1} (\frac{7}{4}) + \cot^{- 1} (\frac{19}{4}) + \cot^{- 1} (\frac{39}{4}) + \cot^{- 1} (\frac{67}{4}) + . . .$ is: [2025]

Smart Achievers · Accepted Answer

(2)

The given infinite series is :

$\cot^{- 1} (\frac{7}{4}) + \cot^{- 1} (\frac{19}{4}) + \cot^{- 1} (\frac{39}{4}) + \cot^{- 1} (\frac{67}{4}) + . . .$

$\Rightarrow T_{n} = \cot^{- 1} (\frac{4 n^{2} + 3}{4})$

$= \tan^{- 1} (\frac{4}{4 n^{2} + 3}) = \tan^{- 1} (\frac{1}{n^{2} + \frac{3}{4}})$

$= \tan^{- 1} (\frac{1}{n^{2} - \frac{1}{4} + 1}) = \tan^{- 1} (\frac{(n + \frac{1}{2}) - (n - \frac{1}{2})}{1 + (n + \frac{1}{2}) (n - \frac{1}{2})})$

$= \tan^{- 1} (n + \frac{1}{2}) - \tan^{- 1} (n - \frac{1}{2})$

$T_{1} = \tan^{- 1} \frac{3}{2} - \tan^{- 1} \frac{1}{2}$

$T_{2} = \tan^{- 1} \frac{5}{2} - \tan^{- 1} \frac{3}{2}$ and so on.

$∴ \sum_{r = 1}^{n} T_{r} = \tan^{- 1} (n + \frac{1}{2}) - \tan^{- 1} \frac{1}{2}$

$as n \to \infty$

$\sum_{r = 1}^{\infty} T_{r} = \frac{π}{2} - \tan^{- 1} \frac{1}{2}$

Solution

Q.
The sum of the infinite series $\cot^{- 1} (\frac{7}{4}) + \cot^{- 1} (\frac{19}{4}) + \cot^{- 1} (\frac{39}{4}) + \cot^{- 1} (\frac{67}{4}) + . . .$ is: [2025]

1 $\frac{π}{2} + \cot^{- 1} (\frac{1}{2})$

2 $\frac{π}{2} - \tan^{- 1} (\frac{1}{2})$

3 $\frac{π}{2} + \tan^{- 1} (\frac{1}{2})$

4 $\frac{π}{2} - \cot^{- 1} (\frac{1}{2})$

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Solution

Q. The sum of the infinite series cot–1(74)+cot–1(194)+cot–1(394)+cot–1(674)+... is: [2025]

1 π2+cot–1(12)

2 π2–tan–1(12)

3 π2+tan–1(12)