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++
// 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
// 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
# 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#
// 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 // 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 ?> |
Javascript
<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++
// 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
// 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
# 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#
// 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 // 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 ?> |
Javascript
<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)