limn→∞ ∑k=1n tan−1⁡12k2 is - Infinity Learn by Sri Chaitanya

A
$π / 8$
B
$π / 2$
C
$π / 4$
D
$π / 3$

Solution:

$\begin{array}{l} \tan^{- 1} (\frac{1}{2 k^{2}}) = \tan^{- 1} \frac{2}{1 + (2 k - 1) (2 k + 1)} \\ = \tan^{- 1} \frac{(2 k + 1) - (2 k - 1)}{1 + (2 k - 1) (2 k + 1)} \\ = \tan^{- 1} (2 k + 1) - \tan^{- 1} (2 k - 1) \end{array}$

$\begin{array}{l} \sum_{k = 1}^{n} \tan^{- 1} (\frac{1}{2 k^{2}}) = \tan^{- 1} (2 n + 1) - \tan^{- 1} (2 n - 1) + \tan^{- 1} \\ (2 n - 1) - \tan^{- 1} (2 n - 3) + \dots + \tan^{- 1} \\ 3 - \tan^{- 1} 1 \\ = \tan^{- 1} (2 n + 1) - \tan^{- 1} 1 \\ \begin{aligned} lim_{n \to \infty} \sum_{k = 1}^{n} \tan^{- 1} (\frac{1}{2 k^{2}}) = \tan^{- 1} \infty - \frac{π}{4} & = \frac{π}{2} - \frac{π}{4} = \frac{π}{4} \end{aligned} \end{array}$

$lim_{n \to \infty} \sum_{k = 1}^{n} \tan^{- 1} (\frac{1}{2 k^{2}})$ is

Solution:

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