Suma de series 2/3 – 4/5 + 6/7 – 8/9 + ——- hasta n términos

Dado el valor de n, encuentra la suma de la serie (2/3) – (4/5) + (6/7) – (8/9) + – – – – – – – hasta n términos.
Ejemplos: 
 

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++ 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 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 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# 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 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.
?>

<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>

Producción : 

0.744012

Complejidad de tiempo: O(n), donde n representa el entero dado.
Espacio auxiliar: O(1), no se requiere espacio adicional, por lo que es una constante.