# Program to find sum of harmonic series

Harmonic series is inverse of a arithmetic progression. In general, the terms in a harmonic progression can be denoted as 1/a, 1/(a + d), 1/(a + 2d), 1/(a + 3d) …. 1/(a + nd).
As Nth term of AP is given as ( a + (n – 1)d). Hence, Nth term of harmonic progression is reciprocal of Nth term of AP, which is 1/(a + (n – 1)d), where “a” is the 1st term of AP and “d” is a common difference.

Method #1: Simple approach

## C

 `// C program to find sum of harmonic series ` `#include ` ` `  `// Function to return sum of harmonic series ` `double` `sum(``int` `n) ` `{ ` `  ``double` `i, s = 0.0; ` `  ``for` `(i = 1; i <= n; i++) ` `      ``s = s + 1/i; ` `  ``return` `s; ` `} ` ` `  `int` `main() ` `{ ` `    ``int` `n = 5; ` `    ``printf``(``"Sum is %f"``, sum(n)); ` `    ``return` `0; ` `}`

## Java

 `// Java Program to find sum of harmonic series ` `import` `java.io.*; ` ` `  `class` `GFG { ` `     `  `    ``// Function to return sum of ` `    ``// harmonic series ` `    ``static` `double` `sum(``int` `n) ` `    ``{ ` `      ``double` `i, s = ``0.0``; ` `      ``for` `(i = ``1``; i <= n; i++) ` `          ``s = s + ``1``/i; ` `      ``return` `s; ` `    ``} ` `  `  `    `  `    ``// Driven Program ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``5``; ` `        ``System.out.printf(``"Sum is %f"``, sum(n));         ` `    ``} ` `} `

## Python3

 `# Python program to find the sum of harmonic series ` ` `  `def` `sum``(n): ` `    ``i ``=` `1` `    ``s ``=` `0.0` `    ``for` `i ``in` `range``(``1``, n``+``1``): ` `        ``s ``=` `s ``+` `1``/``i; ` `    ``return` `s; ` ` `  `# Driver Code  ` `n ``=` `5` `print``(``"Sum is"``, ``round``(``sum``(n), ``6``)) `

## C#

 `// C# Program to find sum of harmonic series ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Function to return sum of ` `    ``// harmonic series ` `    ``static` `float` `sum(``int` `n) ` `    ``{ ` `        ``double` `i, s = 0.0; ` `         `  `        ``for` `(i = 1; i <= n; i++) ` `            ``s = s + 1/i; ` `             `  `        ``return` `(``float``)s; ` `    ``} ` ` `  `     `  `    ``// Driven Program ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 5;         ` `        ``Console.WriteLine(``"Sum is "` `                           ``+ sum(n));         ` `    ``} ` `} `

## PHP

 ` `

Output:

```Sum is 2.283333
```

Method #2: Using recursion

## C++

 `// CPP program to find sum of  ` `// harmonic series using recursion  ` `#include ` `using` `namespace` `std; ` ` `  `float` `sum(``float` `n)  ` `{  ` `    ``// Base condition  ` `    ``if` `(n < 2)  ` `        ``return` `1;  ` ` `  `    ``else` `        ``return` `1 / n + (sum(n - 1));  ` `}  ` ` `  `// Driven Code  ` `int` `main()  ` `{ ` `    ``cout << (sum(8)) << endl;  ` `    ``cout << (sum(10)) << endl;  ` `    ``return` `0; ` `}  ` ` `  `// This code is contributed by ` `// Shashank_Sharma `

## Java

 `// Java program to find sum of  ` `// harmonic series using recursion  ` `import` `java.io.*;  ` ` `  `class` `GFG  ` `{  ` ` `  `float` `sum(``float` `n)  ` `{  ` `    ``// Base condition  ` `    ``if` `(n < ``2``)  ` `        ``return` `1``;  ` ` `  `    ``else` `        ``return` `1` `/ n + (sum(n - ``1``));  ` `}  ` ` `  `// Driven Code  ` `public` `static` `void` `main(String args[])  ` `{  ` `  ``GFG g = ``new` `GFG();  ` `  ``System.out.println(g.sum(``8``));  ` `  ``System.out.print(g.sum(``10``));  ` `}  ` `}  ` ` `  `// This code is contributed by Shivi_Aggarwal  `

## Python 3

 `# Python program to find sum of ` `# harmonic series using recursion ` ` `  `def` `sum``(n): ` ` `  `    ``# Base condition ` `    ``if` `n < ``2``: ` `        ``return` `1` ` `  `    ``else``: ` `        ``return` `1` `/` `n ``+` `(``sum``(n ``-` `1``)) ` `         `  `print``(``sum``(``8``)) ` `print``(``sum``(``10``)) `

## C#

 `//C# program to find sum of  ` `// harmonic series using recursion  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` ` `  `static` `float` `sum(``float` `n)  ` `{  ` `    ``// Base condition  ` `    ``if` `(n < 2)  ` `        ``return` `1;  ` ` `  `    ``else` `        ``return` `1 / n + (sum(n - 1));  ` `}  ` ` `  `// Driven Code  ` `public` `static` `void` `Main()  ` `{  ` `    ``Console.WriteLine(sum(8));  ` `    ``Console.WriteLine(sum(10));  ` `}  ` `}  ` ` `  `// This code is contributed by shs.. `

## PHP

 ` `

Output:

```2.7178571428571425
2.9289682539682538
```

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