∑n=0∞(n!)22n+1(2n+1)!=∑n=0∞∫10tn(1?t)n2n+1dt=2∫10∑0∞tn(1?t)n2ndt=∫101(t?12)+14dt=2 arctan2(t?12)|10=π
特为http://www.cnblogs.com/zhangwenbiao/p/3705281.html写的代码.
时间: 2025-01-07 09:50:22
∑n=0∞(n!)22n+1(2n+1)!=∑n=0∞∫10tn(1?t)n2n+1dt=2∫10∑0∞tn(1?t)n2ndt=∫101(t?12)+14dt=2 arctan2(t?12)|10=π
特为http://www.cnblogs.com/zhangwenbiao/p/3705281.html写的代码.