Determine the value of ∑k=1narctan(12k2)\sum_{k=1}^{n} \arctan\left(\frac{1}{2k^2}\right)∑k=1narctan(2k21) as n→∞n \to \inftyn→∞.
π4\frac{\pi}{4}4π
π2\frac{\pi}{2}2π
π3\frac{\pi}{3}3π
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