Sum of the Series 1 + x/1 + x^2/2 + x^3/3 + .. + x^n/n

This is a mathematical series program where a user must enter the number of terms up to which the sum of the series is to be found. Following this, we also need the value of x, which forms the base of the series.

Examples :

Input : base = 2, range = 5
Output : 18.07

Input : base = 1, range = 10
Output : 3.93

Method 1 (Simple) We just need to follow the series and put the values of the base at x and value range at n and get the sum.

C++

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// C++ program to find sum of series
// 1 + x/1 + x^2/2 + x^3/3 + ....+ x^n/n
#include <math.h>
#include <iostream>
#include <boost/format.hpp>
class gfg
{
public :
double sum(int x, int n)
{
    double i, total = 1.0;
    for (i = 1; i <= n; i++)
        total = total + 
                (pow(x, i) / i);
    return total;
}
};
// Driver code
int main()
{
    gfg g;
    int x = 2;
    int n = 5;
    //std::cout<<g.sum(x,n);
    std::cout << boost::format("%.2f") % g.sum(x,n);
    return 0;
}

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C

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// C program to find sum of series
// 1 + x/1 + x^2/2 + x^3/3 + ....+ x^n/n
#include <math.h>
#include <stdio.h>
  
double sum(int x, int n)
{
    double i, total = 1.0;
    for (i = 1; i <= n; i++)
        total = total + 
                (pow(x, i) / i);
    return total;
}
  
// Driver code
int main()
{
    int x = 2;
    int n = 5;
    printf("%.2f", sum(x, n));
    return 0;
}

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Java

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// Java program to find sum of series
// 1 + 1/x + x^2/2 + x^3/3 + ....+ x^n/n
import static java.lang.Math.pow;
  
class GFG
{
      
// Java code to print the 
// sum of the series
static double sum(int x, int n)
{
    double i, total = 1.0;
    for (i = 1; i <= n; i++)
        total = total + 
                (Math.pow(x, i) / i);
  
    return total;
}
  
// Driver code
public static void main(String[] args)
{
    int x = 2;
    int n = 5;
    System.out.printf("%.2f", sum(x, n));
}
}
  
// This code is contributed by
// Smitha Dinesh Semwal

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Python3

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# Python3 code to find sum of series
# 1 + x/1 + x^2/2 + x^3/3 + .. .+ x^n/n
  
def SUM(x, n):
    total = 1
    for i in range(1, n + 1):
        total = total + ((x**i)/i)
    return total 
  
# Driver Code
x = 2
n = 5
s = SUM(x, n)
print(round(s, 2))

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C#

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// C# program to find sum of series
// 1 + 1/x + x^2/2 + x^3/3 + ....+ x^n/n
using System;
  
class GFG 
{
  
    // Java code to print the
    // sum of the series
    static float sum(int x, int n)
    {
        double i, total = 1.0;
        for (i = 1; i <= n; i++)
            total = total + 
                    (Math.Pow(x, i) / i);
  
        return (float)total;
    }
  
    // Driver code
    public static void Main()
    {
        int x = 2;
        int n = 5;
        Console.WriteLine(sum(x, n));
    }
}
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP program to find sum of series
// 1 + x/1 + x^2/2 + x^3/3 + ....+ x^n/n
  
// Code to print the sum 
// of the series
function sum($x, $n)
{
    $i; $total = 1.0;
    for ($i = 1; $i <= $n; $i++)
        $total = $total
                 (pow($x, $i) / $i);
    return $total;
}
  
// Driver code
$x = 2;
$n = 5;
echo(sum($x, $n));
  
// This code is contributed by Ajit.
?>

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Output :

18.07

Method 2 (Optimized) We can avoid use of pow() function and reuse the previously computed power.

C

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// C program to find sum of series 
// 1 + x^2/2 + x^3/3 + ....+ x^n/n
#include <math.h>
#include <stdio.h>
  
// C code to print the sum
// of the series
double sum(int x, int n)
{
    double i, total = 1.0, multi = x;
    for (i = 1; i <= n; i++) {
        total = total + multi / i;
        multi = multi * x;
    }
    return total;
}
  
// Driver code
int main()
{
    int x = 2;
    int n = 5;
    printf("%.2f", sum(x, n));
    return 0;
}

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Java

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// Java program to find sum of series 
// 1 + x^2/2 + x^3/3 + ....+ x^n/n
  
class GFG
{
  
// Java code to print the sum 
// of the given series
static double sum(int x, int n)
{
    double i, total = 1.0, multi = x;
    for (i = 1; i <= n; i++) 
    {
        total = total + multi / i;
        multi = multi * x;
    }
    return total;
}
  
// Driver code
public static void main(String[] args)
{
    int x = 2;
    int n = 5;
    System.out.printf("%.2f", sum(x, n));
}
}
  
// This code is contributed by
// Smitha Dinesh Semwal

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Python3

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# Python 3 program to find sum of series 
# 1 + x^2/2 + x^3/3 + ....+ x^n/n
  
# Python 3 code to print the 
# sum of the series
def sum(x, n):
  
    total = 1.0
    multi =
    for i in range(1, n+1): 
        total = total + multi /
        multi = multi *
      
    return total 
  
  
# Driver code
x = 2
n = 5
print(round(sum(x, n), 2)) 
  
# This code is contributed by
# Smitha Dinesh Semwal

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C#

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// C# program to find sum of series
// 1 + x^2/2 + x^3/3 + ....+ x^n/n
using System;
  
class GFG
{
  
    // Java code to print the sum
    // of the given series
    static float sum(int x, int n)
    {
        double i, total = 1.0, multi = x;
        for (i = 1; i <= n; i++) 
        {
            total = total + multi / i;
            multi = multi * x;
        }
        return (float)total;
    }
  
    // Driver code
    public static void Main()
    {
        int x = 2;
        int n = 5;
        Console.WriteLine(sum(x, n));
    }
}
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP program to find sum of series 
// 1 + x^2/2 + x^3/3 + ....+ x^n/n
  
// code to print the sum
// of the series
function sum($x, $n)
{
    $i; $total = 1.0; $multi = $x;
    for ($i = 1; $i <= $n; $i++) 
    {
        $total = $total + $multi / $i;
        $multi = $multi * $x;
    }
    return $total;
}
  
// Driver code
$x = 2;
$n = 5;
echo(sum($x, $n));
  
// This code is contributed by Ajit.
?>

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Output :

18.07


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