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^{}. $$.
Proof.
There are several infinite series representations of $e$. In this proof, we will use the most common one, the Taylor expansion^{} of $e$:
$\sum _{i=0}^{\mathrm{\infty}}}{\displaystyle \frac{1}{i!}}={\displaystyle \frac{1}{0!}}+{\displaystyle \frac{1}{1!}}+{\displaystyle \frac{1}{2!}}+\mathrm{\cdots}+{\displaystyle \frac{1}{n!}}+\mathrm{\cdots}.$  (1) 
We chop up the Taylor expansion of $e$ into two parts: the first part $a$ consists of the sum of the first two terms, and the second part $b$ consists of the sum of the rest, or $ea$. The proof of the proposition now lies in the estimation of $a$ and $b$.
Step 1: e$\mathrm{>}$2. First, $a=\frac{1}{0!}+\frac{1}{1!}=1+1=2$. Next, $b>0$, being a sum of the terms in (1), all of which are positive (note also that $b$ must be bounded because (1) is a convergent series^{}). Therefore, $e=a+b=2+b>2+0=2$.
Step 2: e$$3. This step is the same as showing that $$. With this in mind, let us compare term by term of the series (2) representing $b$ and another series (3):
$\frac{1}{2!}}+{\displaystyle \frac{1}{3!}}+\mathrm{\cdots}+{\displaystyle \frac{1}{n!}}+\mathrm{\cdots$  (2) 
and
$\frac{1}{{2}^{21}}}+{\displaystyle \frac{1}{{2}^{31}}}+\mathrm{\cdots}+{\displaystyle \frac{1}{{2}^{n1}}}+\mathrm{\cdots}.$  (3) 
It is wellknown that the second series (a geometric series^{}) sums to 1. Because both series are convergent, the termbyterm comparisons make sense. Except for the first term, where $\frac{1}{2!}=\frac{1}{2}=\frac{1}{{2}^{21}}$, we have $$ for all other terms. The inequality^{} $$, for $n$ a positive number can be translated into the basic inequality $n!>{2}^{n1}$, the proof of which, based on mathematical induction, can be found here (http://planetmath.org/AnExampleOfMathematicalInduction).
Because the term comparisons show

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

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that at least one term from (2) $$ than the corresponding term from (3),
we conclude that (2) $$ (3), or that $$. This concludes the proof. ∎
Title  proof that e is not a natural number 

Canonical name  ProofThatEIsNotANaturalNumber 
Date of creation  20130322 15:39:52 
Last modified on  20130322 15:39:52 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  10 
Author  CWoo (3771) 
Entry type  Proof 
Classification  msc 40A25 
Classification  msc 40A05 
Classification  msc 11J72 
Related topic  EIsTranscendental 