Exponential of Sum/Real Numbers/Proof 3

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


 * $\exp x = \displaystyle \lim_{n \to +\infty} \left({1 + \frac x n}\right)^n$

Note that from Powers of Group Elements we can presuppose the exponent combination laws for natural number indices.

By definition:

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


 * $\displaystyle \left({1 + \frac {x + y} n + \frac {x y} {n^2} }\right)^n \to \left({1 + \frac {x + y} n}\right)^n$ as $n \to +\infty$

To formalize this claim:


 * $\exp \left({x + y}\right) = \exp x \cdot \exp y \iff \dfrac {\exp x \cdot \exp y} {\exp \left({x + y}\right)} = 1$

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:


 * $\displaystyle \sum_{k \mathop = 1}^n \, \left\vert{\binom n k n^{-k} \left({\frac {x y} {n + x + y} }\right)^k }\right\vert$

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

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


 * $\displaystyle 0 \le \sum_{k \mathop = 1}^n \, \left\vert{ {n \choose k} n^{-k} \left({ \frac {xy} {n + x + y} }\right)^k }\right\vert \le \sum_{k \mathop = 1}^n \left\vert{ \frac {xy} {n + x + y} }\right\vert^k$

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


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

This means:


 * $\dfrac{\left({1 + \dfrac{x + y} n + \dfrac {x y} {n^2} }\right)^n} {\left({1 + \dfrac {x + y} n}\right)^n} \to 1$ as $n \to +\infty$

which is equivalent to our hypothesis:


 * $\left({1 + \dfrac{x + y} n + \dfrac {x y} {n^2} }\right)^n \to \left({1 + \dfrac {x + y} n}\right)^n$ as $n \to +\infty$