Sum of the series 1 / 1 + (1 + 2) / (1 * 2) + (1 + 2 + 3) / (1 * 2 * 3) + – – – – – – upto n terms.

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Given a positive integer n, find the sum of the series upto n terms.
Examples :

Input : n = 4 
Output : 3.91667
Series : (1 / 1) + (1 + 2) / (1 * 2) + (1 + 2 + 3) / (1 * 2 * 3) + (1 + 2 + 3 + 4) / (1 * 2 * 3 * 4)
       = 1 / 1 + 3 / 2 + 6 / 6 + 10 / 24
       = 3.91667

Input : n = 6
Output : 4.07083
Series : (1 / 1) + (1 + 2) / (1 * 2) + (1 + 2 + 3) / (1 * 2 * 3)  + (1 + 2 + 3 + 4) / (1 * 2 * 3 * 4) + (1 + 2 + 3 + 4 + 5) / (1 * 2 * 3 * 4 * 5)  + (1 + 2 + 3 + 4 + 5 + 6) / (1 * 2 * 3 * 4 * 5 * 6)
       = 1 / 1 + 3 / 2 + 6 / 6 + 10 / 24 + 15 / 120 + 21 / 720
       = 4.07083

C++

// CPP program to find sum of series. 
#include <bits/stdc++.h> 
using namespace std; 
  
double sumOfSeries(int n) 
{ 
    double res = 0.0 ; 
    int sum = 0, prod = 1; 
    for (int i = 1 ; i <= n ; i++) 
    { 
        sum += i; 
        prod *= i; 
        res += ((double)sum / prod); 
    } 
    return res; 
} 
  
// Driver Code 
int main() 
{ 
    int n = 4 ; 
    cout << sumOfSeries(n) ;     
    return 0; 
} 

Java

// Java program to find sum of series. 
class GFG  
{ 
static double sumOfSeries(int n)  
{ 
    double res = 0.0; 
    int sum = 0, prod = 1; 
    for (int i = 1; i <= n; i++) { 
    sum += i; 
    prod *= i; 
    res += ((double)sum / prod); 
    } 
    return res; 
} 
  
// Driver code 
public static void main(String arg[]) { 
    int n = 4; 
    System.out.println(sumOfSeries(n)); 
} 
} 
  
// This code is contributed by Anant Agarwal. 

Python3

# Python program to 
# find sum of series. 
  
def sumOfSeries(n) : 
  
    res = 0.0
    sum = 0
    prod = 1
    for i in range(1, n + 1) : 
      
        sum = sum + i 
        prod = prod * i 
        res = res + (sum / prod) 
          
    return res 
      
# Driver Code 
n = 4
print (round(sumOfSeries(n), 5))  
  
# This code is contributed by  
# Manish Shaw(manishshaw1) 

C#

// C# program to find sum of series. 
using System; 
  
class GFG { 
  
    static double sumOfSeries(int n) 
    { 
        double res = 0.0; 
        int sum = 0, prod = 1; 
        for (int i = 1; i <= n; i++) { 
            sum += i; 
            prod *= i; 
            res += ((double)sum / prod); 
        } 
        return res; 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int n = 4; 
        Console.Write(sumOfSeries(n)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program to find sum of series. 
  
function sumOfSeries($n) 
{ 
    $res = 0.0 ; 
    $sum = 0; $prod = 1; 
    for ($i = 1 ; $i <= $n ; $i++) 
    { 
        $sum += $i; 
        $prod *= $i; 
        $res += ((double)$sum / $prod); 
    } 
    return $res; 
} 
  
// Driver Code 
$n = 4 ; 
echo(sumOfSeries($n)) ;  
  
// This code is contributed by Ajit. 
?> 

Javascript

<script> 
  
// javascript program to find sum of series. 
  
function sumOfSeries(n) 
{ 
    let res = 0.0 ; 
    let sum = 0, prod = 1; 
    for (let i = 1 ; i <= n ; i++) 
    { 
        sum += i; 
        prod *= i; 
        res += (sum / prod); 
    } 
    return res; 
} 
  
// Driver Code 
let n = 4 ; 
   document.write(sumOfSeries(n).toFixed(5)) ;   
     
// This code contributed by aashish1995  
  
</script>

Output :

3.91667

Time complexity: O(n) since using a single loop

Auxiliary Space: O(1) for constant space for variables

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