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

Last Updated : 11 Aug, 2022

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```

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

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## Javascript

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Output:

`131`

Time complexity: O(sqrt(n)) as for loop will run by sqrt(n) times

Auxiliary Space: O(1)
Note: If observed closely, we can see that, if we take n common, series turns into an Harmonic Progression.