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limn→∞∑m=1ntan−111+m+m2=?\large \lim_{n \to \infty} \sum_{m=1}^n \tan^{-1} \frac 1{1+m+m^2} = ?n→∞limm=1∑ntan−11+m+m21=?
Note: tan−1x∈(−π2,π2)\tan^{-1} x \in \left(-\dfrac \pi 2, \dfrac \pi 2 \right)tan−1x∈(−2π,2π)
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