Equivalence of Definitions of Complex Exponential Function/Power Series Expansion equivalent to Limit of Sequence

Proof
Let:
 * $\ds s_n = \sum_{k \mathop = 0}^n \dfrac {z^k} {k!}$
 * $a_n = \paren {1 + \dfrac z n}^n$

Then we can express $a_n$ as follows:

The limit of the difference between the $k$th terms of $a_n$ and $s_n$ is:

To show that $s_n$ and $a_n$ have the same limit, let $\epsilon \in \R_{>0}$.

From Tail of Convergent Series tends to Zero, it follows that we can find $M \in \N$ such that for all $m \ge M$:


 * $\ds \sum_{k \mathop = m}^n \cmod {\dfrac {z^k} {k!} } < \dfrac \epsilon 2$

For all $k \in \left\{ {0, 1, \ldots, M - 1}\right\}$, we can find $N_k \in \N$ such that for all $n \ge N_k$:


 * $\ds \cmod {\dfrac {z^k} {k!} - \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } < \dfrac \epsilon {2 M}$

Then for all $n \ge \max \paren {M, N_0, N_1, \ldots, N_{M - 1} }$, we have:

As an Absolutely Convergent Series is Convergent, $\sequence {s_n}$ converges.

Then:

The result follows.