3 2tarctan-=arctan+ arctan arctan 18 3 18 -1 假设sA1= arctan Sk arctan + arctan 2k2-arctan k+1 s arctan → arctan1 π(n→∞) n+1 故∑ arctan n=1 2n2418 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