// C语言s=1/n+1/(n+1)+1/(n+2)+…+1/m之和.其中:n<=m
#include
using namespace std;
double fn2m(int n,int m)
{
double x = 0;
for(int i=n;i
{
x += 1.0 / double(i);
}
return x;
}
double p(int n,int x)
{
if(n == 0)
return 1;
if(n == 1)
return x;
return (2*n-1)*x*p(n-1,x) - (n-1)*p(n-2,x)/n;
}
int main()
{
cout }