Sum = \(\frac { 1 }{ 1 }+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{n}\)
#include<stdio.h>
#include<conio.h>
void main ()
{
int n, i;
float sum;
printf (“Enter the number of terms\n”);
scanf (“%. d” , & n);
i = 1;.
do
{
sum = sum + 1 /i;
i = i + 1;
}
while (i <= n);
printf (“% f\ sum);
getch ();
}
