Unizor - Limits - Number e as a Limit

We have introduced a number e, an extremely important number in calculus and analysis, as a base of an exponential function y=a^x with a steepness of 1 at the argument value x=0. We have also indicated without sufficient rigorousness that this number is the limit of an infinite sequence:
e = lim [n→∞] (1+1/n)^n.
In this lecture we will rigorously prove that the above limit does exist and can be used as a definition of a number e.

In the lecture Problems 3 of this chapter about limits we have proved that an infinite monotonously increasing sequence that is bounded from above has a limit. This theorem is a foundation of this lecture. We will prove that a sequence (1+1/n)^n is monotonously increasing and has an upper bound. Therefore, it has a limit as n→∞, and that limit is a definition of a number e.

The fact that our sequence (1+1/n)^n is bounded from above has already been proven in the previous lecture about a function F(x)=e^x where we have proved that for any natural number n
2 ≤ (1+1/n)n ≤ 3
So, all we have to prove now is the monotonic character of this sequence.

So, let's prove that for all n greater than 1
(1+1/n)^n ≥ [1+1/(n−1)]^(n−1)
Direct method to prove this is to use the Binomial formula by Newton.
The expression on the left is a sum of n positive terms, the ith of them being
n!/[(n−i)!·i!·n^i]
The expression on the right is a sum of n−1 positive terms, the i-th of them being
(n−1)!/[(n−i−1)!·i!·(n−1)^i]
The expression on the left has one more positive term in the sum. Now we will prove that every i-th term of the left expression is greater than the corresponding i-th term of the right expression.
Indeed, we can get rid of i! since it is the same for both left and right common terms.
What remains from the left term can be written as
n·(n−1)·...·(n−i+1)/n^i =
= 1·[1−1/n]·...·[1−(i−1)/n]
What remains from the right term can be written as
(n−1)·(n−2)·...·(n−i)/(n−1)^i =
= 1·[1−1/(n−1)]·...·[1−(i−1)/(n−1)]
The term on the left is greater because each member of a product in the left term is greater than corresponding member of a product in the right term since
1-k/n ≥ 1-k/(n−1)
for each k from 1 to i−1.

This completes the proof that a sequence (1+1/n)^n is monotonically increasing. Together with the fact that it is less than 3 for any positive integer number n, proven in the previous lecture about a function ex (that is, 3 is the upper bound for this sequence) it proves that a sequence (1+1/n)^n tends to a limit, which is some real number between 2 and 3. This number is the number e introduced in the previous lecture from a different angle, as a base of an exponential function that has a steepness of 1 at point x=0.

Based on this, let's mention one more equality related to limits.
Let x be any real number. Consider an exponential function F(x)=e^x.
We can state now that
lim[n→∞](1+x/n)^n = e^x

Intuitively, it's obvious since
(1+x/n)^n = [(1+x/n)^(n/x)]^ x
When n goes to infinity, n/x goes to infinity as well and the expression in square brackets tends to e, so the whole expression tents to e^x. Although it's not an absolutely rigorous proof, this consideration should suffice for now.