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# Program to find the sum of the series (1/a + 2/a^2 + 3/a^3 + … + n/a^n)

Given two integers a and n. The task is to find the sum of the series 1/a + 2/a2 + 3/a3 + … + n/an.
Examples:

Input: a = 3, n = 3
Output: 0.6666667
The series is 1/3 + 1/9 + 1/27 which is
equal to 0.6666667
Input: a = 5, n = 4
Output: 0.31039998

Approach: Run a loop from 1 to n and get the term of the series by calculating term = (i / ai). Sum all the generated terms which is the final answer.
Below is the implementation of the above approach:

## C++

 // C++ program to find the sum of// the given series#include #include #include  using namespace std; // Function to return the// sum of the seriesfloat getSum(int a, int n){    // variable to store the answer    float sum = 0;    for (int i = 1; i <= n; ++i)    {         // Math.pow(x, y) returns x^y        sum += (i / pow(a, i));    }    return sum;} // Driver codeint main(){    int a = 3, n = 3;         // Print the sum of the series    cout << (getSum(a, n));    return 0;} // This code is contributed// by Sach_Code

## Java

 // Java program to find the sum of the given seriesimport java.util.Scanner; public class HelloWorld {     // Function to return the sum of the series    public static float getSum(int a, int n)    {        // variable to store the answer        float sum = 0;        for (int i = 1; i <= n; ++i) {             // Math.pow(x, y) returns x^y            sum += (i / Math.pow(a, i));        }        return sum;    }     // Driver code    public static void main(String[] args)    {        int a = 3, n = 3;         // Print the sum of the series        System.out.println(getSum(a, n));    }}

## Python 3

 # Python 3 program to find the sum of# the given seriesimport math # Function to return the# sum of the seriesdef getSum(a, n):     # variable to store the answer    sum = 0;    for i in range (1, n + 1):             # Math.pow(x, y) returns x^y        sum += (i / math.pow(a, i));             return sum; # Driver codea = 3; n = 3;     # Print the sum of the seriesprint(getSum(a, n)); # This code is contributed# by Akanksha Rai

## C#

 // C# program to find the sum// of the given seriesusing System; class GFG{     // Function to return the sum// of the seriespublic static double getSum(int a, int n){    // variable to store the answer    double sum = 0;    for (int i = 1; i <= n; ++i)    {         // Math.pow(x, y) returns x^y        sum += (i / Math.Pow(a, i));    }    return sum;} // Driver codestatic public void Main (){    int a = 3, n = 3;     // Print the sum of the series    Console.WriteLine(getSum(a, n));}} // This code is contributed by jit_t



## Javascript



Output

0.666667

Time Complexity: O(nlogn)
Auxiliary Space: O(1)

Method: Finding the sum of series without using pow function

## C++

 #include #includeusing namespace std; int main(){   // sum of series  int n = 3, a = 3;  double s = 0;   // iterating for loop n times  for (int i = 1; i < n + 1; i++)  {     // finding sum    s = s + (i / (pow(a, i)));  }   // printing the result  cout << s;  return 0;} // This code is contributed by ksrikanth0498.

## Java

 /*package whatever //do not write package name here */// Java code to print// sum of seriesimport java.io.*; class GFG {         public static void main (String[] args) {        int n = 3,a = 3,s = 0;        //iterating for loop n times        for (int i = 1; i <= n; i++) {            //finding sum            s = s + (i/(a**i))                  }        //printing the result        System.out.println(s);    }}// This code is contributed by Atul_kumar_Shrivastava

## Python3

 # Python code to print# sum of seriesn = 3; a = 3; s = 0 # iterating for loop n timesfor i in range(1, n + 1):     # finding sum  s = s + (i/(a**i))   # printing the resultprint(s) # this code is contributed by Gangarajula Laxmi

## C#

 using System;public class GFG{     static public void Main ()    {         // sum of series        int n = 3, a = 3, s = 0;               // iterating for loop n times        for (int i = 1; i < n + 1; i++)        {                       // finding sum            s = s + (i / (Math.Pow(a, i)));        }               // printing the result        Console.WriteLine(s);     } //   this code is contributed by Gangarajula Laxmi

## Javascript



Output

0.666667

Time complexity: O(nlogn) since a single using for loop and logn for inbuilt pow() function.
Auxiliary Space: O(1)

### Method 3: Using the formula for the sum of a geometric series

1. The series 1/a + 2/a^2 + 3/a^3 + … + n/a^n is a geometric series with first term 1/a and common ratio 1/a. The sum of a geometric series is given by the formula:
S = a(1 – r^n)/(1 – r)

where S is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms. Substituting the values for this series, we get:

S = (1/a)(1 – (1/a)^n)/(1 – 1/a)

## C++

 #include #include  using namespace std; double sum_of_series(double a, int n) {    return (1/a)*(1 - pow((1/a),n))/(1 - 1/a);} int main() {    cout << sum_of_series(4, 5) << endl;    return 0;}

## Java

 public class Main {  public static double sumOfSeries(double a, int n) {    return (1/a)*(1 - Math.pow((1/a), n))/(1 - 1/a);  }   public static void main(String[] args) {    System.out.println(sumOfSeries(4, 5));  }}

## Python3

 def sum_of_series(a, n):    return (1/a)*(1 - (1/a)**n)/(1 - 1/a) # Example usage:print(sum_of_series(4, 5))

## C#

 using System; public class Program{    public static double SumOfSeries(int a, int n)    {        return (1.0/a)*(1 - Math.Pow(1.0/a, n))/(1 - 1.0/a);    }         public static void Main()    {        Console.WriteLine(SumOfSeries(4, 5));    }}

## Javascript

 function sum_of_series(a, n) {    return (1/a)*(1 - Math.pow((1/a),n))/(1 - 1/a);} console.log(sum_of_series(4, 5));

Output

0.3330078125

Time complexity: O(n)
Auxiliary Space: O(n)