Proof. There are several infinite series representations of $e$ . In this proof, we will use the most common one, the Taylor expansion of $e$ : \begin{eqnarray} \sum_{i=0}^{\infty}\frac{1}{i!}=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots+\frac{1}{n!}+\cdots. \end{eqnarray} 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): \begin{eqnarray} \frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}+\cdots \end{eqnarray}and \begin{eqnarray} \frac{1}{2^{2-1}}+\frac{1}{2^{3-1}}+\cdots+\frac{1}{2^{n-1}}+\cdots. \end{eqnarray}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.

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 $b<1$ . This concludes the proof.