3 3=s,tartan o=arctan +arctan=arctan s 假设s -1 k-1=arctan -1 k Sk= arctan+arctan=arctan / 2k +1 n S= arctan n n+1 → arctan=(n→>∞) 故∑ arctan 必S4 4 上页18 1 arctan 3 2 = arctan + 18 1 s3 = s2 + arctan , 4 3 = arctan arctan1 1 arctan → +  = n n sn ( ) 4 →   = n . 2 4 1 arctan 1 2   =  n= n 故 , 1 1 arctan k k sk − 假设 − = 2 2 1 arctan 1 arctan k k k sk + − = , 1 arctan + = k k