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

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Theorem

The following definitions of the concept of Complex Exponential Function are equivalent:

As a Power Series Expansion

The exponential function can be defined as a (complex) power series:

\(\ds \forall z \in \C: \, \) \(\ds \exp z\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}\)
\(\ds \) \(=\) \(\ds 1 + \frac z {1!} + \frac {z^2} {2!} + \frac {z^3} {3!} + \cdots + \frac {z^n} {n!} + \cdots\)

As a Limit of a Sequence

The exponential function can be defined as a limit of a sequence:

$\ds \exp z := \lim_{n \mathop \to \infty} \paren {1 + \dfrac z n}^n$


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:

\(\ds a_n\) \(=\) \(\ds \sum_{k \mathop = 0}^n \binom n k \paren {\dfrac z n}^k 1^{n - k}\) Binomial Theorem: Integral Index
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n \dfrac {z^k} {k!} \dfrac {n \paren {n - 1} \cdots \paren {n - k + 1} }{n^k}\)
\(\ds \) \(=\) \(\ds 1 + \sum_{k \mathop = 1}^n \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n}\) by algebraic manipulations


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

\(\ds \lim_{n \mathop \to +\infty} \cmod {\dfrac {z^k} {k!} - \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} }\) \(=\) \(\ds \cmod {\lim_{n \mathop \to +\infty} \paren {\dfrac {z^k} {k!} - \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } }\) Modulus of Limit
\(\ds \) \(=\) \(\ds \cmod {\dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \lim_{n \mathop \to +\infty} \dfrac j n} } }\) Combination Theorem for Limits of Complex Functions
\(\ds \) \(=\) \(\ds \cmod {\dfrac {z^k} {k!} \paren {1 - 1} }\) Sequence of Powers of Reciprocals is Null Sequence
\(\ds \) \(=\) \(\ds 0\)


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:

\(\ds \cmod {s_n - a_n}\) \(=\) \(\ds \cmod {\sum_{k \mathop = 1}^{M - 1} \dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } + \sum_{k \mathop = M}^n \dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } }\)
\(\ds \) \(\le\) \(\ds \sum_{k \mathop = 1}^{M - 1} \cmod {\dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k- 1 } \paren {1 - \dfrac j n} } } + \sum_{k \mathop = M}^n \cmod {\dfrac {z^k} {k!} } \cmod {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} }\) Triangle Inequality for Complex Numbers
\(\ds \) \(\le\) \(\ds \sum_{k \mathop = 1}^{M - 1} \cmod {\dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } } + \sum_{k \mathop = M}^n \cmod {\dfrac {z^k} {k!} }\)
\(\ds \) \(<\) \(\ds \sum_{k \mathop = 1}^{M - 1} \dfrac \epsilon {2 M} + \dfrac \epsilon 2\)
\(\ds \) \(=\) \(\ds \epsilon\)

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

Then:

\(\ds 0\) \(=\) \(\ds \lim_{n \mathop \to +\infty} \cmod {s_n - a_n}\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to +\infty} s_n - \lim_{n \mathop \to +\infty} a_n\) Combination Theorem for Limits of Complex Functions
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {z^k} {k!} - \lim_{n \mathop \to +\infty} \paren {1 + \dfrac z n}^n\)

The result follows.

$\blacksquare$

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