Exponential of Sum/Real Numbers/Proof 3

Lemma
This proof assumes the definition of $\exp$ as defined by a limit of a sequence:


 * $\exp x = \displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n$

From Powers of Group Elements we can presuppose the Exponent Combination Laws for natural number indices.

By definition:

Intuitively, the $\paren {1 + \dfrac {x + y} n}$ term is the most influential of the terms involved in the limit, and:


 * $\paren {1 + \dfrac {x + y} n + \dfrac {x y} {n^2} }^n \to \paren {1 + \dfrac {x + y} n}^n$ as $n \to +\infty$

To formalize this claim:


 * $\map \exp {x + y} = \exp x \cdot \exp y \iff \dfrac {\exp x \cdot \exp y} {\map \exp {x + y} } = 1$

Now, as $n \to +\infty$, we use the Combination Theorem for Limits of Functions to investigate the behavior of this sequence, term by term.

As $1$ trivially converges to $1$, consider now the other terms of the sequence.

We invoke the Squeeze Theorem for Absolutely Convergent Series.

Hence it will suffice to investigate the limit behaviour of:


 * $\displaystyle \sum_{k \mathop = 1}^n \, \size {\binom n k n^{-k} \paren {\frac {x y} {n + x + y} }^k}$

From $\dbinom n k$ is not greater than $n^k$:

Therefore, we may conclude, using Absolute Value is Bounded Below by Zero:


 * $\displaystyle 0 \le \sum_{k \mathop = 1}^n \size { {n \choose k} n^{-k} \paren {\frac {x y} {n + x + y} }^k} \le \sum_{k \mathop = 1}^n \size {\frac {x y} {n + x + y} }^k$

From Sum of Infinite Geometric Progression, the right hand term converges to:


 * $0 \to 0$ as $n \to +\infty$, trivially.

This means:


 * $\dfrac {\paren {1 + \dfrac {x + y} n + \dfrac {x y} {n^2} }^n} {\paren {1 + \dfrac {x + y} n}^n} \to 1$ as $n \to +\infty$

which is equivalent to our hypothesis:


 * $\paren {1 + \dfrac {x + y} n + \dfrac {x y} {n^2} }^n \to \paren {1 + \dfrac {x + y} n}^n$ as $n \to +\infty$