Question

to calculate sum of the series :(1/1!)+(2/2!)+(3/3!)+(4/4!)+............+(n/n!)

Answer 1

#include<stdio.h>
#include<conio.h>
int main()
{
int n;
float sum = 0,d,fact =1,j,i;
clrscr();
printf("Enter the number:");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
fact=1;
for (j=1;j<=i;j++)
{
fact=fact*j;
}
d=(float)i/(float)fact;
sum=sum+d;
}
printf("sum = %.2f", sum);
getch();
return 0;
}

Answer 2

#include <stdio.h>

int main(){
    const int N=10000;
   
    double sum = 0;
    double fac = 1; // 0! = 1
    for(int n=1; n<=N; n++){
        fac *= n;
        sum += n/fac;
    }
   
    printf("1/1!+2/2!+3/3!+...+N/N! = %f ; (with N=%d)", sum, N);
    return 0;
}

Comments

N/N! = 1/(N-1)!                  ... this would be better