# Sum of series (n/1) + (n/2) + (n/3) + (n/4) +…….+ (n/n)

Given a value n, find the sum of series, (n/1) + (n/2) + (n/3) + (n/4) +…….+(n/n) where the value of n can be up to 10^12.
Note: Consider only integer division.

Examples:

```Input : n = 5
Output : (5/1) + (5/2) + (5/3) +
(5/4) + (5/5) = 5 + 2 + 1 + 1 + 1
= 10

Input : 7
Output : (7/1) + (7/2) + (7/3) + (7/4) +
(7/5) + (7/6) + (7/7)
= 7 + 3 + 2 + 1 + 1 + 1 + 1
= 16
```

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

Below is the program to find the sum of given series:

## C++

 `// CPP program to find ` `// sum of given series ` `#include ` `using` `namespace` `std; ` ` `  `// function to find sum of series ` `long` `long` `int` `sum(``long` `long` `int` `n) ` `{ ` `    ``long` `long` `int` `root = ``sqrt``(n); ` `    ``long` `long` `int` `ans = 0; ` ` `  `    ``for` `(``int` `i = 1; i <= root; i++)  ` `        ``ans += n / i; ` `     `  `    ``ans = 2 * ans - (root * root); ` `    ``return` `ans; ` `} ` ` `  `// driver code ` `int` `main() ` `{ ` `    ``long` `long` `int` `n = 35; ` `    ``cout << sum(n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find ` `// sum of given series ` `import` `java.util.*; ` ` `  `class` `GFG { ` `     `  `    ``// function to find sum of series ` `    ``static` `long` `sum(``long` `n) ` `    ``{ ` `        ``long` `root = (``long``)Math.sqrt(n); ` `        ``long` `ans = ``0``; ` `      `  `        ``for` `(``int` `i = ``1``; i <= root; i++)  ` `            ``ans += n / i; ` `          `  `        ``ans = ``2` `* ans - (root * root); ` `         `  `        ``return` `ans; ` `    ``} ` `     `  `    ``/* Driver code */` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``long` `n = ``35``; ` `        ``System.out.println(sum(n)); ` `    ``} ` `} ` `     `  `// This code is contributed by Arnav Kr. Mandal.         `

## Python3

 `# Python 3 program to find ` `# sum of given series ` ` `  `import` `math ` ` `  `# function to find sum of series ` `def` `sum``(n) : ` `    ``root ``=` `(``int``)(math.sqrt(n)) ` `    ``ans ``=` `0` `  `  `    ``for` `i ``in` `range``(``1``, root ``+` `1``) : ` `        ``ans ``=` `ans ``+` `n ``/``/` `i ` `      `  `    ``ans ``=` `2` `*` `ans ``-` `(root ``*` `root) ` `    ``return` `ans ` ` `  `# driver code ` `n ``=` `35` `print``(``sum``(n)) ` ` `  `# This code is contributed by Nikita Tiwari. `

## C#

 `// C# program to find  ` `// sum of given series ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Function to find sum of series ` `    ``static` `long` `sum(``long` `n) ` `    ``{ ` `        ``long` `root = (``long``)Math.Sqrt(n); ` `        ``long` `ans = 0; ` `     `  `        ``for` `(``int` `i = 1; i <= root; i++)  ` `            ``ans += n / i; ` `         `  `        ``ans = 2 * ans - (root * root); ` `         `  `        ``return` `ans; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``long` `n = 35; ` `        ``Console.Write(sum(n)); ` `    ``} ` `} ` `     `  `// This code is contributed vt_m.  `

## PHP

 ` `

Output:

`131`

Note: If observed closely, we can see that, if we take n common, series turns into an Harmonic Progression.

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