Factorial of a number simply returns out the multiplication of that number with all numbers lesser than the number up to 1 over which factorial is applied. Now the question arises of whether the series is convergent or divergent. According to concepts of Infinite series and factorial in mathematics, the series may not converge but it can contain a convergent subsequence.
Illustration:
n! = n * (n-1) × (n-2) × (n-3) × (n-4) × …… × 4 × 3 × 2 × 1
Input : n = 5
Processing : 1/1! + 2/2! + 3/3! + 4/4! + 5/5!
1 + 2/2 + 3/6 + 4/24 + 5/120
Output : 2.708333333333333
Approach :
- Enter the number of terms N.
- Create a function to calculate the sum of series, say, calculateSum.
- In calculateSum function(), create a variable sum which stores the total sum of the series.
- Run a loop N times.
- Call factorial function() to calculate the factorial of a given number.
- Return sum.
Implementation:
Java
// Java Program to Find Sum of the Series// 1/1! + 2/2! + 3/3! + ……1/N!// Importing java generic librariesimport java.io.*;class GFG { // Function(recursive) to calculate factorial public static double factorial(int i) { // Step1: Base case if (i == 1) { return 1; } // Step2&3: Recursion execution & call statements return i * factorial(i - 1); } // Function to calculate sum of series public static double calculateSum(int N) { // Store total_sum in sum double sum = 0; // Iteration by running a loop N times for (int i = 1; i <= N; i++) { sum = sum + ((double)i / factorial(i)); } // Return calculated final sum return sum; } // Main driver method public static void main(String[] args) { /* No of terms in series taken inn order to show output */ int N = 5; // Print sum of series by // calling function calculating sum of series System.out.println("The sum of series upto " + N + " terms is : " + calculateSum(N)); }} |
Output
The sum of series upto 5 terms is : 2.708333333333333
Time Complexity: O(n)
Auxiliary Space: O(n) for call stack
