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