Limits...

Calculus Level 3

lim n m = 1 n tan 1 1 1 + m + m 2 = ? \large \lim_{n \to \infty} \sum_{m=1}^n \tan^{-1} \frac 1{1+m+m^2} = ?

Note : tan 1 x ( π 2 , π 2 ) \tan^{-1} x \in \left(-\dfrac \pi 2, \dfrac \pi 2 \right)

π 2 \frac \pi 2 π 4 \frac \pi 4 None of the others Does not exist

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1 solution

Trishit Chandra
Mar 4, 2015

Yep exactly the same as I did.

Kartik Sharma - 6 years, 3 months ago

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