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Sum of series 2/3 - 4/5 + 6/7 - 8/9 + ------- upto n terms

Last Updated : 20 Feb, 2023
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Given the value of n, find the sum of the series (2 / 3) - (4 / 5) + (6 / 7) - (8 / 9) + - - - - - - - upto n terms.
Examples : 
 

Input : n = 5
Output : 0.744012
Series : (2 / 3) - (4 / 5) + (6 / 7) - (8 / 9) + (10 / 11)

Input : n = 7
Output : 0.754268
Series : (2 / 3) - (4 / 5) + (6 / 7) - (8 / 9) +
         (10 / 11) - (12 / 13) + (14 / 15)


 


 

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

// Function to find sum of series
// up-to n terms
double seriesSum(int n)
{
    // initializing counter by 1
    int i = 1;
    
    // variable to calculate result
    double res = 0.0;
    bool sign = true;
    
    // while loop until nth term 
    // is not reached
    while (n > 0) 
    {
        n--;
        
        // boolean type variable 
        // for checking validation
        if (sign) {
            sign = !sign;
            res = res + (double)++i / ++i;
        }
        else {
            sign = !sign;
            res = res - (double)++i / ++i;
        }
    }
    
    return res;
}

// Driver Code
int main()
{
    int n = 5;
    cout << seriesSum(n);    
    return 0;
}
Java 
// Java program to find 
// sum of given series
import java.io.*;

class GFG {
    
    // Function to find sum of series
    // up-to n terms
    static double seriesSum(int n)
    {
    
    // initializing counter by 1
    int i = 1;
    
    // variable to calculate result
    double res = 0.0;
    boolean sign = true;
    
    // while loop until nth term 
    // is not reached
    while (n > 0) 
    {
        n--;
        
        // boolean type variable 
        // for checking validation
        if (sign)
        {
            sign = !sign;
            res = res + (double)++i / ++i;
        }
        
        else 
        {
            sign = !sign;
            res = res - (double)++i / ++i;
        }
    }
    
    return res;
}
    
    // Driver Code
    public static void main (String[] args) {
        
        int n = 5;
        
        System.out.print(seriesSum(n));
    }
}

// This code is contributed by vt_m
Python3 
# Python3 program to find 
# sum of given series

# Function to find
# sum of series
# up-to n terms
def seriesSum(n):
    
    # initializing 
    # counter by 1
    i = 1;
    
    # variable to 
    # calculate result
    res = 0.0;
    sign = True;
    
    # while loop until nth 
    # term is not reached
    while (n > 0): 
        n = n - 1;
        
        # boolean type variable 
        # for checking validation
        if (sign):
            sign = False;
            res = res + (i + 1) / (i + 2);
            i = i + 2;
        else:
            sign = True;
            res = res - (i + 1) / (i + 2);
            i = i + 2;
    
    return res;

# Driver Code
n = 5;
print(round(seriesSum(n), 6)); 

# This code is contributed 
# by mits
C# 
// C# program to find 
// sum of given series
using System;

class GFG {
    
    // Function to find sum of 
    // series up-to n terms
    static double seriesSum(int n)
    {
    
    // initializing counter by 1
    int i = 1;
    
    // variable to calculate result
    double res = 0.0;
    bool sign = true;
    
    // while loop until nth term 
    // is not reached
    while (n > 0) 
    {
        n--;
        
        // boolean type variable 
        // for checking validation
        if (sign)
        {
            sign = !sign;
            res = res + (double)++i / ++i;
        }
        
        else
        {
            sign = !sign;
            res = res - (double)++i / ++i;
        }
    }
    
    return res;
}
    
    // Driver Code
    public static void Main () {
        
        int n = 5;
        Console.Write(seriesSum(n));
    }
}

// This code is contributed by vt_m
PHP 
<?php
// PHP program to find 
// sum of given series

// Function to find sum of series
// up-to n terms
function seriesSum($n)
{
    // initializing counter by 1
    $i = 1;
    
    // variable to calculate result
    $res = 0.0;
    $sign = true;
    
    // while loop until nth term 
    // is not reached
    while ($n > 0) 
    {
        $n--;
        
        // boolean type variable 
        // for checking validation
        if ($sign) {
            $sign = !$sign;
            $res = $res + (double)++$i / ++$i;
        }
        else {
            $sign = !$sign;
            $res = $res - (double)++$i / ++$i;
        }
    }
    
    return $res;
}

// Driver Code
$n = 5;
echo(seriesSum($n)); 

// This code is contributed by Ajit.
?>
JavaScript 
<script>

// javascript program to find 
// sum of given series

// Function to find sum of series
// up-to n terms
function seriesSum( n)
{
    // initializing counter by 1
    let i = 1;
    
    // variable to calculate result
    let res = 0.0;
    let sign = true;
    
    // while loop until nth term 
    // is not reached
    while (n > 0) 
    {
        n--;
        
        // boolean type variable 
        // for checking validation
        if (sign) {
            sign = !sign;
            res = res + ++i / ++i;
        }
        else {
            sign = !sign;
            res = res - ++i / ++i;
        }
    }
    
    return res;
}
// Driver Code
let n = 5 ;
   document.write(seriesSum(n).toFixed(6)) ;
   
// This code contributed by aashish1995

</script>

Output : 

0.744012

Time Complexity: O(n), where n represents the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.


N

nikunj_agarwal
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