例題2-3　近似計算

#include<stdio.h> 
int main() 
{  
	double sum = 0; 
	for(int i = 0; ; i++) 
	{    
		double term = 1.0 / (i*2+1);    
		if(i % 2 == 0) 		sum += term;    
		else sum -= term;   if(term < 1e-6) break;  
	}  
	printf("%.6f\n", sum);
  	return 0; 
}