proof that e is not a natural number

Here, we are going to show that the natural log base $e$ is not a natural number by showing a sharper result: that $e$ is between $2$ and $3$.

Proposition. .

Proof.

There are several infinite series representations of $e$. In this proof, we will use the most common one, the Taylor expansion of $e$:

${\displaystyle \sum _{i=0}^{\mathrm{\infty }}}{\displaystyle \frac{1}{i!}}={\displaystyle \frac{1}{0!}}+{\displaystyle \frac{1}{1!}}+{\displaystyle \frac{1}{2!}}+\mathrm{\cdots }+{\displaystyle \frac{1}{n!}}+\mathrm{\cdots }.$(1)

We chop up the Taylor expansion of $e$ into two parts: the first part $a$ consists of the sum of the first two terms, and the second part $b$ consists of the sum of the rest, or $e-a$. The proof of the proposition now lies in the estimation of $a$ and $b$.

Step 1: e$\mathrm{>}$2. First, $a=\frac{1}{0!}+\frac{1}{1!}=1+1=2$. Next, $b>0$, being a sum of the terms in (1), all of which are positive (note also that $b$ must be bounded because (1) is a convergent series). Therefore, $e=a+b=2+b>2+0=2$.

Step 2: e3. This step is the same as showing that . With this in mind, let us compare term by term of the series (2) representing $b$ and another series (3):

${\displaystyle \frac{1}{2!}}+{\displaystyle \frac{1}{3!}}+\mathrm{\cdots }+{\displaystyle \frac{1}{n!}}+\mathrm{\cdots }$(2)

and

${\displaystyle \frac{1}{2^{2-1}}}+{\displaystyle \frac{1}{2^{3-1}}}+\mathrm{\cdots }+{\displaystyle \frac{1}{2^{n-1}}}+\mathrm{\cdots }.$(3)

It is well-known that the second series (a geometric series) sums to 1. Because both series are convergent, the term-by-term comparisons make sense. Except for the first term, where $\frac{1}{2!}=\frac{1}{2}=\frac{1}{2^{2-1}}$, we have for all other terms. The inequality , for $n$ a positive number can be translated into the basic inequality $n!>2^{n-1}$, the proof of which, based on mathematical induction, can be found here (http://planetmath.org/AnExampleOfMathematicalInduction).

Because the term comparisons show

• that the terms from (2) $\le$ the corresponding terms from (3), and

• that at least one term from (2) than the corresponding term from (3),

we conclude that (2) (3), or that . This concludes the proof.∎