Equivalence of Definitions of Real Exponential Function/Limit of Sequence implies Sum of Series

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Theorem

The following definition of the concept of the real exponential function:

As the Limit of a Sequence

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

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


implies the following definition:

As the Sum of a Series

The exponential function can be defined as a power series:

$\exp x := \displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$


Proof

Let $\exp x$ be the real function defined as the limit of the sequence:

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

From the General Binomial Theorem:

\(\displaystyle \paren {1 + \frac x n}^n\) \(=\) \(\displaystyle 1 + x + \frac {n \paren {n - 1} x^2} {2! \ n^2} + \frac {n \paren {n - 1} \paren {n - 2} x^3} {3! \ n^3} + \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^0} {0!} + \frac {x^1} {1!} + \paren {\frac {n - 1} n} \frac {x^2} {2!} + \paren {\frac {\paren {n - 1} \paren {n - 2} } {n^2} } \frac {x^3} {3!} + \cdots\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0\) \(=\) \(\displaystyle \paren {1 + \frac x n}^n - \frac {x^0} {0!} + \frac {x^1} {1!} + \paren {\frac {n - 1} n} \frac {x^2} {2!} + \paren {\frac {\paren {n - 1} \paren {n - 2} } {n^2} } \frac {x^3} {3!} + \cdots\)

From Power over Factorial, this converges to:

$\exp x - \dfrac {x^0} {0!} + \dfrac {x^1} {1!} + \dfrac {x^2} {2!} + \dfrac {x^3} {3!} + \cdots = 0$

as $n \to +\infty$:

\(\displaystyle \leadsto \ \ \) \(\displaystyle \exp x\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\)

$\blacksquare$