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/*********************************************
Approximating e by series expansion

Using the formula:

   e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + ...

Write a program that takes a number of terms
from the user and prints out the approximation
of e given by that number of terms
*********************************************/
#include <stdio.h>

int factorial(int n);

/*********************************************
 ** main() function
 *********************************************/
int main() {
  // Get n, the number of terms, from user
  int n;
  printf("How many terms in the series expansion? ");
  fflush(stdout);
  scanf(" %d", &n);

  // Compute and sum n terms of series
  double e = 0;
  for(int i=0; i < n; i++) {
    e += 1.0/factorial(i);
  }

  // Print approximation to screen
  printf("e is approximately %.20f\n", e);

  return 0;
}

/*********************************************
 ** factorial - Does what you think!
 *********************************************/
int factorial(int n) {
  int ans = 1;
  for(int i = 2; i <= n; i++) {
    ans = ans*i;
  }
  return ans;
}