# Exponential Function is Well-Defined/Real/Proof 2

## 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.

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

$\blacksquare$