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Evaluate ∑m=1∞tan−1(8m16m4−32m2+5).\large \sum_{m=1}^{\infty} \tan^{-1} \left(\frac{8m}{16m^4-32m^2+5}\right).m=1∑∞tan−1(16m4−32m2+58m).
Remember that the range of the inverse tangent function is (−π2,π2) ( - \frac{ \pi}{2} , \frac{ \pi }{2}) (−2π,2π).
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