Sum of the Series 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + . . . . .

Given a positive integer n, the problem is to find the sum of the given series of n terms: 
1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + . . . . . . . + 1/(n*(n+1)) 

Examples:  

Input : 3
Output : 0.75
  ( 1/(1*2)+ 1/(2*3) + 1/(3*4) )
= (1/2 + 1/6 + 1/12)
= 0.75

Input : 10
Output : 0.909
  ( 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) +
   1/(5*6) + 1/(6*7) + 1/(7*8) + 1/(8*9) +
   1/(9*10) + 1/(10*11) )
= (1/2 + 1/6 + 1/12 + 1/20 + 1/30 +
   1/42 + 1/56 + 1/72 + 1/90 + 1/110)
= 0.909 

Naive Approach: Use a for loop to calculate each term iteratively and add it to the final sum. 

// C++ program to find the sum of given series
#include <bits/stdc++.h>
using namespace std;

// function to find the sum of given series
double sumOfTheSeries(int n)
{
    // Computing sum term by term
    double sum = 0.0;
    for (int i = 1; i <= n; i++)
        sum += 1.0 / (i * (i + 1));   
    return sum;
}

// driver program to test above function
int main()
{
    int n = 10;
    cout << sumOfTheSeries(n);
    return 0;
}

// Java program to find the sum of given series
class demo {

    // function to find the sum of given series
    public static double sumOfTheSeries(int n)
    {
       // Computing sum term by term
        double sum = 0.0;
        for (int i = 1; i <= n; i++) 
            sum += 1.0 / (i * (i + 1));
        return sum;
    }

    // driver program to test above function
    public static void main(String args[])
    {
        int n = 10;
        System.out.println(sumOfTheSeries(n));
    }
}

# Python3 code to find the sum of given series

# Function to find the sum of given series
def sumOfTheSeries( n ):
    
    # Computing sum term by term
    sum = 0
    for i in range(1, n + 1):
        sum += 1.0 / (i * (i + 1)); 
    return sum


# Driver function
if __name__ == '__main__':
    
    ans = sumOfTheSeries(10)
    
    # Rounding decimal value to 6th decimal place
    print (round(ans, 6))

# This code is contributed by 'saloni1297'

// C# program to find the sum of given series
using System;

class demo {

    // Function to find the sum of given series
    public static double sumOfTheSeries(int n)
    {
        // Computing sum term by term
        double sum = 0.0;
        for (int i = 1; i <= n; i++) 
            sum += 1.0 / (i * (i + 1));
        return sum;
    }

    // Driver Code
    public static void Main()
    {
        int n = 10;
        Console.Write(sumOfTheSeries(n));
    }
}

// This code is contributed by vt_m

<?php
// PHP program to find the
// sum of given series

// function to find the 
// sum of given series
function sumOfTheSeries( $n)
{
    // Computing sum term by term
    $sum = 0.0;
    for ( $i = 1; $i <= $n; $i++)
        $sum += 1.0 / ($i * ($i + 1)); 
    return $sum;
}

// Driver Code
$n = 10;
echo sumOfTheSeries($n);

// This code is contributed by anuj_67
?>

<script>

// JavaScript program to find the sum of given series

    // function to find the sum of given series
    function sumOfTheSeries(n)
    {
       // Computing sum term by term
        let sum = 0.0;
        for (let i = 1; i <= n; i++) 
            sum += 1.0 / (i * (i + 1));
        return sum;
    }

// Driver code    
         
        let n = 10;
        document.write(sumOfTheSeries(n));
            
</script>

Output : 

0.909091

Efficient Approach: Use the formula n/(n+1)  

Validity of the formula:
Sum upto n terms = 1/(1*2) + 1/(2*3) + 1/(3*4) +
                           ........ + 1/(n*(n+1))
where
1st term = 1/(1*2)
2nd term = 1/(2*3)
3rd term = 1/(3*4)
.
.
.
.
n-th term = 1/(n*(n+1))

i.e. the k-th term is of the form 1/(k*(k+1))
which can further be written as k-th term = 
                              1/k - 1/(k+1)

So sum upto n terms can be calculated as:
  (1/1 - 1/1+1) + (1/2 - 1/2+1) + (1/3 - 1/3+1)
   + ......... + (1/n-1 - /1n) + (1/n - 1/n+1) 
= (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + .........
                    + (1/n-1 - 1/n) + (1/n - 1/n+1)
= 1 - 1/n+1
= ((n+1) - 1)/n+1
= n/n+1

Hence sum  upto n terms = n/n+1

// C++ program to find sum of given series
#include <bits/stdc++.h>
using namespace std;

// function to find sum of given series
double sumOfTheSeries(int n)
{
    // type-casting n/n+1 from int to double
    return (double)n / (n + 1);
}

// driver program to test above function
int main()
{
    int n = 10;
    cout << sumOfTheSeries(n);
    return 0;
}

// Java program to find sum of given series
class demo {

    // function to find sum of given series
    public static double sumOfTheSeries(int n)
    {
        // type -casting n/n+1 from int to double
        return(double)n / (n + 1);
    }

    // driver program to test above function
    public static void main(String args[])
    {
        int n = 10;
        System.out.println(sumOfTheSeries(n));
    }
}

# Python3 code to find sum of given series

# Function to find sum of given series
def sumOfTheSeries(n):
    
    # Type-casting n/n+1 from int to float
    return (float(n) / (n + 1))

# Driver function   
if __name__ == '__main__':
        
    n = 10
    ans = sumOfTheSeries(n)
    
    # Rounding decimal value
    print (round(ans, 6))

# This code is contributed by 'saloni1297'

// C# program to find sum of given series
using System;

class demo {

    // Function to find sum of given series
    public static double sumOfTheSeries(int n)
    {
        // type -casting n/n+1 from int to double
        return(double)n / (n + 1);
    }

    // Driver Code
    public static void Main()
    {
        int n = 10;
        Console.Write(sumOfTheSeries(n));
    }
}

// This code is contributed by vt_m.

<?php
// PHP program to find
// sum of given series

// function to find sum
// of given series
function sumOfTheSeries($n)
{
    // type-casting n/n+1 
    // from int to double
    return $n / ($n + 1);
}

// Driver Code
$n = 10;
echo sumOfTheSeries($n);

// This code is contributed
// by SanjuTomar
?>

<script>
    // Javascript program to find sum of given series
    
    // Function to find sum of given series
    function sumOfTheSeries(n)
    {
        // type -casting n/n+1 from int to double
        return (n / (n + 1));
    }
    
    let n = 10;
      document.write(sumOfTheSeries(n).toFixed(6));
      
</script>

Output :  

0.909091

Time complexity: O(1) as constant operations are performed

Auxiliary space: O(1)