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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 5 2.716666667 6 2.718055556
7 2.718253968
8 2.718278770
9 2.718281526

 1 #include<stdio.h> 
 2 int jc(int n)
 3 {
 4     int i=1,s=1;
 5     if(n==1||n==0)
 6     return 1;
 7     while(i<=n)
 8     {
 9         s*=i;
10
 
         i++;
11     }
12     return s;
13 }
14 int main()
15 {
16     double a[10];
17     a[0]=1;
18     for(int i=1;i<=9;i++)
19     {
20         a[i]=1.0/jc(i)+a[i-1];
21     }
22     printf("n e\n");
23     printf("- -----------\n");
24     for(int i=0;i<=9;i++)
25     {
26         if(i!=8)
27         printf("
 
%d %.10g\n",i,a[i]);
28         else
29         printf("%d %.9f\n",i,a[i]);
30     }  
31     return 0;
32 }

1012 u Calculate e