u Calculate e
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16844 Accepted Submission(s): 7307
where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.
#include<iostream>
using namespace std;
double pow(double i)
{
double sum=1,j;
for(j=1;j<=i;j++)
{
sum*=j;
}
return 1/sum;
}
double he(double i)
{
double sun=0,sum=0,j;
for(j=1;j<=i;j++)
{
sum+=pow(j);
}
return sum;
}
int main()
{
double a,i,j,sum=1,e;
cout<<'n'<<' '<<'e'<<endl;
cout<<'-'<<' '<<"-----------"<<endl;
cout<<'0'<<' '<<'1'<<endl;
cout<<'1'<<' '<<'2'<<endl;
cout<<'2'<<' '<<"2.5"<<endl;
for(i=3;i<=9;i++)
{
cout<<i<<' ';
e=he(i);
printf("%.9f\n",e+1);
}
return 0;
}