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

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

// CPP program to find
// sum of given series
#include <bits/stdc++.h>
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 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.        

# 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# 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
// PHP program to find
// sum of given series

// function to find
// sum of series
function sum($n)
{
    $root = intval(sqrt($n));
    $ans = 0;

    for ($i = 1; $i <= $root; $i++) 
        $ans += intval($n / $i);

    $ans = (2 * $ans) - 
           ($root * $root);
    return $ans;
}

// Driver code
$n = 35;
echo (sum($n));

// This code is contributed by 
// Manish Shaw(manishshaw1)
?>

<script>
// Javascript program to find
// sum of given series

// function to find
// sum of series
function sum(n)
{
    let root = parseInt(Math.sqrt(n));
    let ans = 0;

    for (let i = 1; i <= root; i++)
        ans += parseInt(n / i);

    ans = (2 * ans) -
        (root * root);
    return ans;
}

// Driver code
let n = 35;
document.write(sum(n));

// This code is contributed by gfgking.
</script>

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.