Efficient Program to Compute Sum of Series 1/1! + 1/2! + 1/3! + 1/4! + .. + 1/n!
Given a positive integer n, write a function to compute the sum of the series 1/1! + 1/2! + .. + 1/n!
A Simple Solution is to initialize the sum as 0, then run a loop and call the factorial function inside the loop.
Following is the implementation of a simple solution.
C++
// A simple C++ program to compute sum of series 1/1! + 1/2! + .. + 1/n!#include <iostream>using namespace std;// Utility function to findint factorial(int n){ int res = 1; for (int i=2; i<=n; i++) res *= i; return res;}// A Simple Function to return value of 1/1! + 1/2! + .. + 1/n!double sum(int n){ double sum = 0; for (int i = 1; i <= n; i++) sum += 1.0/factorial(i); return sum;}// Driver program to test above functionsint main(){ int n = 5; cout << sum(n); return 0;} |
Java
// A simple Java program to compute // sum of series 1/1! + 1/2! + .. + 1/n!import java.io.*;class GFG { // Utility function to find static int factorial(int n) { int res = 1; for (int i = 2; i <= n; i++) res *= i; return res; } // A Simple Function to return value // of 1/1! + 1/2! + .. + 1/n! static double sum(int n) { double sum = 0; for (int i = 1; i <= n; i++) sum += 1.0/factorial(i); return sum; } // Driver program public static void main (String[] args) { int n = 5; System.out.println(sum(n)); }}// This code is contributed by Ajit. |
Python3
# Python3 program to compute sum of series# 1/1! + 1/2! + .. + 1/n!# Function to find factorial of a number def factorial(n): res = 1 for i in range(2, n + 1): res *= i return res # A Simple Function to return value # of 1/1! + 1/2! + .. + 1/n!def sum(n): s = 0.0 for i in range(1, n + 1): s += 1.0 / factorial(i) print(s)# Driver program to test above functionsn = 5sum(n)# This code is contributed by Danish Raza |
C#
// A simple C# program to compute sum // of series 1/1! + 1/2! + .. + 1/n!using System;class GFG { // Utility function to find static int factorial(int n) { int res = 1; for (int i = 2; i <= n; i++) res *= i; return res; } // A Simple Function to return value // of 1/1! + 1/2! + .. + 1/n! static double sum(int n) { double sum = 0; for (int i = 1; i <= n; i++) sum += 1.0/factorial(i); return sum; } // Driver program public static void Main () { int n = 5; Console.WriteLine(sum(n)); }}// This code is contributed by Sam007. |
PHP
<?php// A simple PHP program to compute// sum of series 1/1! + 1/2! + .. + 1/n!// Utility function to findfunction factorial($n){ $res = 1; for ($i = 2; $i <= $n; $i++) $res *= $i; return $res;}// A Simple Function to return // value of 1/1! + 1/2! + .. + 1/n!function sum($n){ $sum = 0; for ($i = 1; $i <= $n; $i++) $sum += 1.0 / factorial($i); return $sum;}// Driver Code$n = 5;echo(sum($n));// This code is contributed by Ajit.?> |
Javascript
<script>//Javascript program to compute// sum of series 1/1! + 1/2! + .. + 1/n!// Utility function to findfunction factorial(n){ let res = 1; for (let i = 2; i <= n; i++) res *= i; return res;}// A Simple Function to return// value of 1/1! + 1/2! + .. + 1/n!function sum(n){ let sum = 0; for (let i = 1; i <= n; i++) sum += 1.0 / factorial(i); return sum;}// Driver Codelet n = 5;document.write(sum(n).toFixed(5));// This code is contributed by sravan kumar</script> |
Output:
1.71667
The time complexity of the above solution is O(n * n!) which is huge.
An Efficient Solution can find the sum in O(n) time. The idea is to calculate factorial in the same loop as the sum. Following is the implementation of this idea.
C++
// A simple C++ program to compute sum of series 1/1! + 1/2! + .. + 1/n!#include <iostream>using namespace std;// An Efficient Function to return value of 1/1! + 1/2! + .. + 1/n!double sum(int n){ double sum = 0; int fact = 1; for (int i = 1; i <= n; i++) { fact *= i; // Update factorial sum += 1.0/fact; // Update series sum } return sum;}// Driver program to test above functionsint main(){ int n = 5; cout << sum(n); return 0;} |
Java
// A simple Java program to compute // sum of series 1/1! + 1/2! + .. + 1/n!import java.io.*;class GFG { // An Efficient Function to return // value of 1/1! + 1/2! + .. + 1/n! static double sum(int n) { double sum = 0; int fact = 1; for (int i = 1; i <= n; i++) { // Update factorial fact *= i; // Update series sum sum += 1.0/fact; } return sum; } // Driver program public static void main (String[] args) { int n = 5; System.out.println(sum(n)); }}// This code is contributed by Ajit. |
Python3
# Python3 program to compute sum of series # 1/1! + 1/2! + .. + 1/n!# Function to return value of# 1/1! + 1/2! + .. + 1/n!def sum(n): sum = 0 fact = 1 for i in range(1, n + 1): # Update factorial fact *= i # Update series sum sum += 1.0/fact print(sum)# Driver program to test above functionsn = 5sum(n)# This code is contributed by Danish Raza |
C#
// A simple C# program to compute sum// of series 1/1! + 1/2! + .. + 1/n!using System;class GFG { // An Efficient Function to return // value of 1/1! + 1/2! + .. + 1/n! static double sum(int n) { double sum = 0; int fact = 1; for (int i = 1; i <= n; i++) { // Update factorial fact *= i; // Update series sum sum += 1.0 / fact; } return sum; } // Driver program public static void Main () { int n = 5; Console.WriteLine(sum(n)); }}// This code is contributed by Sam007. |
PHP
<?php// A simple PHP program to // compute sum of series // 1/1! + 1/2! + .. + 1/n!// An Efficient Function to // return value of 1/1! + // 1/2! + .. + 1/n!function sum($n) { $sum = 0; $fact = 1; for ($i = 1; $i <= $n; $i++) { // Update factorial $fact *= $i; // Update series sum $sum += 1.0 / $fact; } return $sum;}// Driver Code$n = 5;echo sum($n);// This code is contributed by vt_m.?> |
Javascript
<script> // A simple Javascript program to compute sum // of series 1/1! + 1/2! + .. + 1/n! // An Efficient Function to return // value of 1/1! + 1/2! + .. + 1/n! function sum(n) { let sum = 0; let fact = 1; for (let i = 1; i <= n; i++) { // Update factorial fact *= i; // Update series sum sum += 1.0 / fact; } return sum.toFixed(5); } let n = 5; document.write(sum(n)); </script> |
Output:
1.71667
This article is contributed by Rahul Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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