The Euler number e

1. The Euler number e

This section is devoted to two beautiful monotonic sequences

$a n = 1 + 1 n n and b n = 1 + 1 n n + 1$

and their limits. Staring at the sequences, it is not clear what will happen: will the $1 + 1 n$ go to 1 so fast that the exponent of $n$ or $n + 1$ not matter? or will the exponent beat the $1 + 1 n$ and the sequence go to infinity? In fact, the answer is something in between. These sequences come up regularly enough that it is worth learning the sequences and the limits, if not the following rather pretty argument.

We start by proving monotonicity. Let $1 ⩽ r ⩽ s$ and apply the exponential inequality to $( 1 + 1 r ) r / s$ giving

$1 + 1 r r s ⩽ 1 + r s 1 r = 1 + 1 s$

so

$1 + 1 r r ⩽ 1 + 1 s s$

showing that $( a n )$ is monotonic nondecreasing.

Now let $1 ⩽ r ⩽ s$ again and consder a second application of the exponential inequality:

$1 - 1 r + 1 r + 1 s + 1 ⩽ 1 - r + 1 s + 1 1 r + 1 = 1 - 1 s + 1$

so

$r r + 1 r + 1 ⩽ s s + 1 s + 1$

whence

$1 + 1 r r + 1 ⩾ 1 + 1 s s + 1 ⁢$

It follows that $( b n )$ is monotonic nonincreasing.

Also $b n = 1 + 1 n a n ⩾ a n$ for all $n$ so given any $k , l ∈ ℕ$ and $n ⩾ k , l$ we have

$a k ⩽ a n ⩽ b n ⩽ b l$

Hence $a k ⩽ b l$ for all $k , l$ and therefore both sequences $( a n )$ and $( b n )$ converge, to $a n → e 1$ and $b n → e 2$, say, where $e 1 ⩽ e 2 ∈ ℝ$. But

$b n - a n = 1 + 1 n n 1 + 1 n - 1 = a n n$

so $b n = a n + a n · 1 n → e 1 + e 1 · 0 = e 1$ by continuity, so $e 1 = e 2$ by uniqueness of limits.

The limit $e$ ( $= e 1 = e 2$ ) that is the limit here can be calculated approximately using the terms of the sequences (though there are better methods that give more accurate answers more quickly) and it turns out that $e = 2.718281828...$, the number commonly (pun intended!) used as the base of the natural log.

Theorem.

The sequences $a n = 1 + 1 n n and b n = 1 + 1 n n + 1$ both converge to $e = 2.718281828...$.

Taking this idea a little further it is possible to define a function $E ( x ) = lim n → ∞ 1 + x n n$. It turns out that $E ( x )$ has all the properties expected of an exponential function, including $E ( 0 ) = 1$, $E ( 1 ) = e$ and $E ( x + y ) = E ( x ) · E ( y )$ and we may define $e x = E ( x )$. Once the inverse function $log$ or $ln$ of $E ( x )$ has been defined we may define $x y = e y log ( x )$ for $x > 0$ and arbitrary $y ∈ ℝ$. But apart from stating the essential ideas here, this takes us far to far for this module, so I shall omit the details.

Exercise.

Modify the arguments at the top of this web page to show that for each $x ∈ ℝ$ the sequence $1 + x n n$ is eventually monotonic nondecreasing and the sequence $1 + x n n + 1$ is eventually monotonic nonincreasing.