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

• Difficulty Level : Medium
• Last Updated : 20 Apr, 2021

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

 ``

## Javascript

 ``

Output:

`131`

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

My Personal Notes arrow_drop_up