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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

 ` `

Output:

```0.6666667
```

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Improved By : jit_t, Sach_Code, Akanksha_Rai

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