1. 用C语言求极限（n到无穷）1+1/2^2+1/3^2+……+1/n^2,近似值为10的负6次方
#include<stdio.h>
const double eps = 1e-6;
int main()
{
double s=0;
double i=1;
while(1.0/(i*i)>=eps)
{
s+=1.0/(i*i);
i+=1.0;
}
printf("%lf\n",s);
return 0;
}
1. 用C语言求极限（n到无穷）1+1/2^2+1/3^2+……+1/n^2,近似值为10的负6次方
#include<stdio.h>
const double eps = 1e-6;
int main()
{
double s=0;
double i=1;
while(1.0/(i*i)>=eps)
{
s+=1.0/(i*i);
i+=1.0;
}
printf("%lf\n",s);
return 0;
}