求e的近似值（误差小于10^-6)

#include
int main()
{float e;long long n;int j,k;for(n=1; k<=1000000; n++){for(j=1,k=1,e=1.0; j<=n; j++){k *= j;e += 1.0/k;}}printf("The value of e is: %f",e);return 0;
}