Editorial

Compute the specified sum using a loop.

Let's calculate the specified sum mathematically:

Now we can use this formula to find the answer.

Algorithm realization

Read the input value of $n$.

scanf("%d",&n);

In the variable $i$, iterate through the numbers $1,2,3,...,n$. On each iteration, add the value of $1/(i\cdot (i+1))$ to the resulting sum $s$.

s = 0; i = 1;
while (i <= n) 
{
  s += 1.0 / i / (i + 1);
  i++;
}

Print the answer.

printf("%lf\n",s);

Algorithm realization — for loop

Read the input value of $n$.

scanf("%d",&n);

Compute the desired sum in the variable $s$.

s = 0;

Using a for loop, compute the sum of $n$ terms. The $i$-th term is equal to $1/(i\cdot (i+1))$.

for (i = 1; i <= n; i++)
  s += 1.0 / (i * (i + 1));

Print the answer.

printf("%.6lf\n",s);

Algorithm realization — formula

Read the input value of $n$.

scanf("%d", &n);

Compute the answer using the formula and print it.

s = 1 - 1.0 / (n + 1);
printf("%.6lf\n", s);

Java realization

import java.util.*;

class Main 
{
  public static void main(String[] args) 
  {
    Scanner con = new Scanner(System.in);
    int n = con.nextInt();

    double s = 0;
    for(int i = 1; i <= n; i++)
      s += 1.0 / (i * (i + 1));

    System.out.printf("%6f\n", s);
    con.close();
  }
}

Python realization

Read the input value of $n$.

n = int(input())

Compute the desired sum in the variable $s$.

s = 0;

Using a for loop, compute the sum of $n$ terms. The $i$-th term is equal to $1/(i\cdot (i+1))$.

for i in range(1, n+1):
  s += 1.0 / (i * (i + 1));

Print the answer.

print(s)