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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. $2<e<3$.
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 $ea$. 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 $b=ea=e2<32=1$. 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
$\displaystyle\frac{1}{2^{{21}}}+\frac{1}{2^{{31}}}+\cdots+\frac{1}{2^{{n1}}% }+\cdots.$  (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 $\frac{1}{n!}<\frac{1}{2^{{n1}}}$ for all other terms. The inequality $\frac{1}{n!}<\frac{1}{2^{{n1}}}$, 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.
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. ∎
Mathematics Subject Classification
40A25 no label found40A05 no label found11J72 no label found Forums
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