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

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

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

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

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

Proof: The lemma is clearly true if

$$1 - \frac{k(k+1)}{2N} \leq 0$$

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

$$1 - \frac{k(k+1)}{2N} \gt 0.$$

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

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

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

so on the one hand we have

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

while by applying the Lemma we have

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

so that

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

Since it is readily established that

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

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

$$\left(1 + \frac{x}{N}\right)^N$$

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