Problem Description
A simple mathematical formula for e is



where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.
 

Output
Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.
 

Sample Output
n e - ----------- 0 1 1 2 2 2.5 3 2.666666667 4 2.708333333
#include<stdio.h>
int main()
{
    double a[11];
    int i,n,b,j;
    a[0]=1.0;
    for(i=1;i<=9;i++)
    {
        b=1;
        for(j=1;j<=i;j++)
        b=b*j;
        a[i]=a[i-1]+(double)1/b;
    }
    printf("n e\n- -----------\n0 1\n1 2\n2 2.5\n");
    for(i=3;i<=9;i++)
    {
        printf("%d %.9lf\n",i,a[i]);
    }
    return 0;
}