# 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 $3$.

###### Proof.

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

 $\displaystyle\sum_{i=0}^{\infty}\frac{1}{i!}=\frac{1}{0!}+\frac{1}{1!}+\frac{1% }{2!}+\cdots+\frac{1}{n!}+\cdots.$ (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$>$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: e$<$3. This step is the same as showing that $b=e-a=e-2<3-2=1$. With this in mind, let us compare term by term of the series (2) representing $b$ and another series (3):

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

and

 $\displaystyle\frac{1}{2^{2-1}}+\frac{1}{2^{3-1}}+\cdots+\frac{1}{2^{n-1}}+\cdots.$ (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 $\frac{1}{n!}<\frac{1}{2^{n-1}}$ for all other terms. The inequality  $\frac{1}{n!}<\frac{1}{2^{n-1}}$, 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) $\leq$ 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 $b<1$. This concludes the proof. ∎

Title proof that e is not a natural number ProofThatEIsNotANaturalNumber 2013-03-22 15:39:52 2013-03-22 15:39:52 CWoo (3771) CWoo (3771) 10 CWoo (3771) Proof msc 40A25 msc 40A05 msc 11J72 EIsTranscendental