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

• Last Updated : 29 Jun, 2022

Given two integers and . 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 series``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 / ``pow``(a, i));``    ``}``    ``return` `sum;``}` `// Driver code``int` `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 series``import` `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 series``import` `math` `# Function to return the``# sum of the series``def` `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 code``a ``=` `3``; n ``=` `3``;``    ` `# Print the sum of the series``print``(getSum(a, n));` `# This code is contributed``# by Akanksha Rai`

## C#

 `// C# program to find the sum``// of the given series``using` `System;` `class` `GFG``{``    ` `// Function to return the sum``// of the series``public` `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 code``static` `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`

## PHP

 ``

## Javascript

 ``

Output:

`0.6666667`

Time Complexity: O(nlogn)

Auxiliary Space: O(1)

Method: Finding the sum of series without using pow function

## Python3

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

Output

`0.6666666666666667`

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

Auxiliary Space: O(1)

My Personal Notes arrow_drop_up