# 1012.u Calculate e

## Problem

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

## Solution

#include<stdio.h>
int main(void)
{
double ji,sum;
int i,j,k;

printf("n e\n");
printf("- -----------\n");

for(i=0;i<=9;i++){
sum=0;
for(j=0;j<=i;j++){
ji=1;
for(k=1;k<=j;k++){
ji=ji*1/k;
}
sum=sum+ji;
}
if(i<=1){
printf("%d %.0f\n",i,sum);
}
else if(i==2){
printf("%d %.1f\n",i,sum);
}
else{
printf("%d %.9f\n",i,sum);
}
}
return 0;
}