근사 공식
\frac{\pi}{\sqrt{12}} = \sum^\infty_{k=0} \frac{(-3)^{-k}}{2k+1}
을 유도해보자. 대학 1학년 미적분학 과정에서 공부하는 멱급수를 이용하여 계산한다.
\frac{1}{1 + t^2} = \sum_{k=0}^\infty \left( -t^2 \right)^{k} \ = \ \sum_{k=0}^\infty (-1)^k t^{2k}
\therefore \quad \int_0^x \frac{1}{1 + t^2} \; dt = \int_0^x \sum_{k=0}^\infty (-1)^k t^{2k} \, dt \ = \ \sum_{k=0}^\infty \int_0^x (-1)^k t^{2k} \, dt
\therefore \quad \arctan x = \sum_{k=0}^\infty (-1)^k \frac{ \ \ x^{2k+1}}{2k + 1} \ = \ x \, \sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{2k + 1}
\therefore \quad \arctan \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \, \sum_{k=0}^\infty (-1)^k \frac{\left( \frac{1}{\sqrt{3}} \right)^{2k}}{2k + 1}
\therefore \quad \frac{\pi}{6} = \frac{1}{\sqrt{3}} \, \sum_{k=0}^\infty (-1)^k \frac{\left( \frac{1}{3} \right)^{k}}{2k + 1}
\therefore \quad \frac{\pi}{\sqrt{12}} = \sum_{k=0}^\infty (-1)^{-k} \frac{3^{-k}}{2k + 1} \ = \ \sum_{k=0}^\infty \frac{(-3)^{-k}}{2k + 1}
