Exponential Function is Well-Defined/Real/Proof 2

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Theorem

Let $x \in \R$ be a real number.

Let $\exp x$ be the exponential of $x$.


Then $\exp x$ is well-defined.


Proof

This proof assumes the sequence definition of $\exp$.

Let $\sequence {f_n}$ be the sequence of mappings $f_n : \R \to \R$ defined as:

$\map {f_n} x = \paren {1 + \dfrac x n}^n$

Fix $x \in \R$.

Then:

\(\displaystyle \map {f_n} x\) \(=\) \(\displaystyle \paren {1 + \dfrac x n}^n\) Definition of $\map {f_n} x$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 0}^n \binom n k \frac {x^k} {n^k}\) Binomial Theorem: Integral Index
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 0}^n \frac {x^k} {k!} \frac {\paren n \times \paren {n - 1} \times \paren {n - 2} \times \cdots \paren {n - k + 1} }{n \times n \times n \times \cdots n}\) Definition of Factorial
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 0}^n \frac {x^k} {k!} \paren 1 \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \cdots \paren {1 - \frac {k - 1} n}\)
\(\displaystyle \) \(\le\) \(\displaystyle \size {\sum_{k \mathop = 0}^n \frac {x^k} {k!} \paren 1 \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \cdots \paren {1 - \frac {k - 1} n} }\) Negative of Absolute Value
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 0}^n \frac {\size x^k} {k!} \paren 1 \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \cdots \paren {1 - \frac {k - 1} n}\) Absolute Value Function is Completely Multiplicative
\(\displaystyle \) \(\le\) \(\displaystyle \sum_{k \mathop = 0}^n \frac {\size x^k} {k!}\) Multiplication of Positive Number by Real Number Greater than One
\(\displaystyle \) \(<\) \(\displaystyle \sum_{k \mathop = 0}^\infty \frac {\size x^k} {k!}\) Sum of positive terms is increasing
\(\displaystyle \) \(<\) \(\displaystyle \infty\) Series of Power over Factorial Converges


Thus, $\sequence {\map {f_n} x}$ is bounded above.

From Exponential Sequence is Eventually Increasing:

$\exists N \in \N: \sequence {\map {f_{N + n} } x}$ is increasing

From Monotone Convergence Theorem (Real Analysis), $\sequence {\map {f_{N + n} } x}$ converges to some $z \in \R$.

From Tail of Convergent Sequence, $\sequence {\map {f_n} x}$ converges to $z$.

Hence the result, from Limit of Function Unique.

$\blacksquare$