The Maclaurin series for arctan(x)\arctan(x)arctan(x) is ∑n=0∞(−1)nx2n+12n+1\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}∑n=0∞2n+1(−1)nx2n+1 for ∣x∣≤1|x| \leq 1∣x∣≤1. Using this, evaluate arctan(1)\arctan(1)arctan(1).
π6\frac{\pi}{6}6π
π4\frac{\pi}{4}4π
π3\frac{\pi}{3}3π
π2\frac{\pi}{2}2π