Замена циклов с помощью рекурсии

function sum(arr, n) {
  if (n <= 0) {
    return 0;
  } else {
    return sum(arr, n - 1) + arr[n - 1];
  }
}

/* 1) sum([1], 0) should equal 0.
   2) sum([2, 3, 4], 1) should equal 2.
   3) sum([2, 3, 4, 5], 3) should equal 9. */

/* My explanations are down below */

/*
    Explanation 1) 
    sum([1], 0) should equal 0.
    n is less or equal to 0 so line 2 works and returns 0 at line 3.
*/

/*  Explanation 2) 
    sum([2, 3, 4], 1) should equal 2
    n is not less or equal to 0 so it will not return 0 according to line 2. We move to line 5.
    *return sum(arr, n - 1) + arr[n - 1];
=>  return sum([2, 3, 4], 1 - 1) + arr[1 - 1];
=>  return sum([2, 3, 4], 0) + arr[0] => n is less or equal to 0 so it will return zero according to line 2.
=>  return 0 + arr[0]
=>  Since arr[0] is equals to 2
=>  return 0 + 2;
=>  2
*/

/*  Explanation 3)
    sum([2, 3, 4, 5], 3) should equal 9
    n is not less or equal to 0 so it will not return 0 according to line 2. We move to line 5.
    *return sum(arr, n - 1) + arr[n - 1];
=>  return sum([2, 3, 4, 5], 3 - 1) + arr[3 - 1];
=>  return sum([2, 3, 4, 5], 2) + arr[2]; => n is 2, not less or equal to 0 so go back to line 5 + arr[2].
=>  return sum(arr, n - 1) + arr[n - 1] + arr[2];
=>  return sum([2, 3, 4, 5], 2 - 1) + arr[2 - 1] + arr[2];
=>  return sum([2, 3, 4, 5], 1) + arr[1] + arr[2]; => n is 1, not less or equal to 0 so goes back to line 5 arr[1] + arr[2].
=>  return sum(arr, n - 1) + arr[n - 1] + arr[1] + arr[2];
=>  return sum([2, 3, 4, 5], 1 - 1) + arr[1 - 1] + arr[1] + arr[2];
=>  return sum([2, 3, 4, 5], 0) + arr [0] + arr[1] + arr[2];
=>  return 0 + arr[0] + arr[1] + arr[2];
=>  in our array, arr[0] is 2, arr[1] is 3, arr[2] is 4.
=>  0 + 2 + 3 + 4
=>  5 + 4
=>  returns 9
*/