programming-examples/c/Series_Programs/C Program to find (1!(by)1) + (2!(by)2) + (3!(by)3) + (4!(by)4) + (5!(by)5) + ...c

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2019-11-15 12:59:38 +01:00
/*(1!/1) + (2!/2) + (3!/3) + (4!/4) + (5!/5) + ... + (n!/n)*/
#include<stdio.h>
long fact(int n)
{
long i, f=1;
for(i=1; i<=n; i++)
{
f=f*i;
}
return f;
}
int main()
{
long i,n;
double sum=0;
printf("Enter value of n ");
scanf("%d",&n);
for(i=1; i<=n; i++)
{
sum=sum+(fact(i)/i);
}
printf("Sum: %lf",sum);
return 0;
}