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Equivalent Defintions of $\displaystyle{e^x}$

It is well-known that for all complex numbers z,

\begin{displaymath}
\lim_{N\rightarrow\infty} \left(1 + \frac{z}{N}\right)^N =
\lim_{N\rightarrow\infty}1 + \sum_{k=1}^N \frac{1}{k!}z^k.\end{displaymath}

What follows below is a direct proof of this fact in the case where z is a positive real number. The proof is elementary in that it does not depend on limits superior and inferior, but instead on the Pinching Theorem.

Lemma 198

Suppose that $\displaystyle{k}$ and $\displaystyle{N}$ are positive integers and $\displaystyle{k \leq N}$.Then

\begin{displaymath}
\left(1-\frac{1}{N}\right)\left(1-\frac{2}{N}\right)\cdots
\left(1-\frac{k}{N}\right) \geq 1 - \frac{k(k+1)}{2N}\end{displaymath}

Proof: The lemma is clearly true if

\begin{displaymath}
1 - \frac{k(k+1)}{2N} \leq 0\end{displaymath}

since the righthand side expression is never negative. Therefore, suppose that

\begin{displaymath}
1 - \frac{k(k+1)}{2N} \gt 0.\end{displaymath}

In this case the proof is by induction on k along with the observation that for $\displaystyle{0 < u, v < 1}$, $\displaystyle{(1-u)(1-v) \gt 1-(u+v)}$. QED

It follows directly from the Binomial Theorem that for $N \geq 2$ and x > 0

\begin{displaymath}
\left(1 + \frac{x}{N}\right)^N 
= 1 + x + \sum_{k=2}^N \frac...
 ...t(1 - \frac{1}{N}\right)\cdots\left(1 - \frac{k-1}{N}\right)x^k\end{displaymath}

so on the one hand we have

\begin{displaymath}
\left(1 + \frac{x}{N}\right)^N 
\leq 1 + x + \sum_{k=2}^N \frac{1}{k!}x^k\end{displaymath}

while by applying the Lemma we have

\begin{displaymath}
\left(1 + \frac{x}{N}\right)^N 
\geq 1 + x + \sum_{k=2}^N \frac{1}{k!}
\left(1 - \frac{k(k-1)}{2N}\right)x^k\end{displaymath}

so that

\begin{displaymath}
\left\vert\left(1 + \frac{x}{N}\right)^N 
- \left(\sum_{k=0}...
 ...t)\right\vert \leq
\frac{1}{2N}\sum_{k=0}^{N-2}\frac{1}{k!}x^k \end{displaymath}

Since it is readily established that

\begin{displaymath}
\lim_{N\rightarrow\infty}\sum_{k=0}^N \frac{1}{k!}x^k\end{displaymath}

exists for positive $\displaystyle{x}$ by using comparison to a geometric series, we see that

\begin{displaymath}
\left(1 + \frac{x}{N}\right)^N\end{displaymath}

has the same limit as $N\rightarrow\infty$.



 

Eric S Key
10/24/2000