Pi 的一种级数表示

求以下级数的和：

\[ S_n = \sum_{n=0}^{\infty} \frac{n!}{(2n+1)!!} \]

解：考虑积分

\[ \int_{0}^{1} (1 - x^2)^n d x = \int_{0}^{\frac{\pi}{2}} \sin^{2n+1} t d t = \frac{(2n)!!}{(2n+1)!!} \]

通项

\[ \frac{n!}{(2n+1)!!} = \frac{1}{2^n} \frac{(2n)!!}{(2n+1)!!} = \frac{1}{2^n} \int_{0}^{1} (1 - x^2)^n d x = \int_{0}^{1} (\frac{1 - x^2}{2})^n d x \]

交换求和与积分的顺序得

\[ \begin{align*} S_n &= \sum_{n=0}^{\infty} \left\{ \int_{0}^{1} (\frac{1 - x^2}{2})^n d x \right\} = \int_{0}^{1} \left\{ \sum_{n=0}^{\infty} (\frac{1 - x^2}{2})^n \right\} d x \\ &= \int_{0}^{1} \frac{2}{1 + x^2} d x = \frac{\pi}{2} \end{align*} \]

另解：利用莱布尼茨级数的欧拉变换可得

\[ \frac{\pi}{4} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \frac{1}{2} \sum_{n=0}^{\infty} \frac{n!}{(2n+1)!!} \]