1. Encountering Euler's Number

2. A random variable question
Suppose values are selected at random from the unit- interval [0,1] until their total exceeds 1 .
Let X count the number of values required.
What is the expected value of X ?
(In other words, what is the mean μX ?)

3. Computer simulation
We have constructed a computer simulation which repeatedly performs this experiment using a random number generating function
It shows the average number of randomly chosen values needed to obtain a total larger than 1

4. Output of 'eulernum.cpp' demo

5. Euler's Number e =
This number e has been called the most important number in mathematics
It turns up often in probability studies
But it was discovered in connection with investigations of compound interest
e = limitn→∞ (1 + 1/n)n

6. Two numerical sequences
Consider the two sets L and U defined by
L = { (1 + 1/n)n | n = 1, 2, 3, ... }
and
U = { (1 + 1/n)n+1 | n = 1, 2, 3, ... }
These set names come from “lower” and “upper”.

7. Three theorems to prove
Numbers in L form an increasing sequence
i.e., (1 + 1/n)n < (1 + 1/(n+1))m+1 , for n > 0
Numbers in U form a decreasing sequence
i.e., (1 + 1/n)n+1 > (1 + 1/(n+1))n+2 , for n > 0
Numbers in L are less than numbers in U
i.e., (1 + 1/n)n < (1 + 1/n)n+1 , for n > 0

8. And this “squeeze” theorem
0 < (1 + 1/n)n+1 – (1 + 1/n)n < 4/n , for n > 0
It implies that the distance which separates numbers in L from numbers in U
Is smaller than any positive real number
(in other words, it's equal to zero)

9. Furthermore... The sum Sn of these reciprocal factorials
Sn = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! /n!
satisfies the inequality
(1 + 1/n)n < Sn < (1 + 1/n)n+1

10. An inequality from algebra
Lemma 1: If 0 ≤ a < b, then
(n+1)an < (bn+1 – an+1) / (b - a) < (n+1)bn
Proof: We use (bn+1 – an+1) = (b - a)( ∑k akbn-k ) and the fact that each of the n+1 summands satisfies
an < akbn-k < bn . ▄

11. Another algebra fact Lemma 2: For any positive integer n , we have
(1 + 1/(n+1))2 < (1 + 1/(n+1)) + 1/n
Proof: Reexpress both sides in terms of their common denominator n(n+1)2 and then observe that their numerators can readily be compared:
n3 + 3n2 + 3n < n3 + 3n2 + 3n + 1 ▄

12. Recall the Binomial Theorem
(1 + 1/n)n = ∑k b(n;k) (1/n)k
(1 + 1/n)n+1 = ∑k b(n+1;k) (1/n)k

13. Timeout for demo and proofs
The foregoing slides lay the groundwork for proving the essential facts about Euler's Number
But we also can utilize a demo-program that shows us the behavior of the successive terms that lie in the sets L and U