Best solutions to - "Find M=∑n=1∞tan−12n2"

Kaydence Huff

Answered question

2022-01-22

Find

M = \sum_{n = 1}^{\infty} \tan^{- 1} \frac{2}{n^{2}}

portafilses

Beginner2022-01-23Added 13 answers

\sum_{n = 1}^{\infty} \tan^{- 1} \frac{2}{n^{2}}

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

= \sum_{n = 1}^{\infty} \tan^{- 1} (1 + n) + \tan^{- 1} (1 - n)

= \sum_{n = 1}^{\infty} \tan^{- 1} (n + 1) - \tan^{- 1} (n - 1)

Edit:

\sum_{n = 1}^{\infty}

was missing in your question before edit. I am not going to delete this.
However your proof is now correct.

M = lim_{m \to \infty} (\tan^{- 1} (m + 1) + \tan^{- 1} m - \tan^{- 1} 1 - \tan^{\tan^{- 1}} 0)

= \frac{π}{2} + \frac{π}{2} - \frac{π}{4} - 0 = π - \frac{π}{4} = \frac{3 π}{4}

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