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

• Difficulty Level : Easy
• Last Updated : 12 Apr, 2021

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```

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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++

 `// C++ program to find sum of series``// 1 + x/1 + x^2/2 + x^3/3 + ....+ x^n/n``#include ``#include ``#include ``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<

## C

 `// C program to find sum of series``// 1 + x/1 + x^2/2 + x^3/3 + ....+ x^n/n``#include ``#include ` `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;``}`

## Java

 `// 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`

## Python3

 `# 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``))`

## C#

 `// 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.`

## PHP

 ``

## Javascript

 ``

Output :

`18.07`

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

## C++

 `// C++ program to find sum of series``// 1 + x^2/2 + x^3/3 + ....+ x^n/n``#include ``using` `namespace` `std;` `// 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;``    ``cout << fixed << setprecision(2) << sum(x, n);``    ``return` `0;``}` `// This code is contributed by shubhamsingh10`

## C

 `// C program to find sum of series``// 1 + x^2/2 + x^3/3 + ....+ x^n/n``#include ``#include ` `// 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;``}`

## Java

 `// 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`

## Python3

 `# 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 ``=` `x``    ``for` `i ``in` `range``(``1``, n``+``1``):``        ``total ``=` `total ``+` `multi ``/` `i``        ``multi ``=` `multi ``*` `x``    ` `    ``return` `total`  `# Driver code``x ``=` `2``n ``=` `5``print``(``round``(``sum``(x, n), ``2``))` `# This code is contributed by``# Smitha Dinesh Semwal`

## C#

 `// 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.`

## PHP

 ``

## Javascript

 ``

Output :

`18.07`

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