Exponential Sequence is Eventually Strictly Positive

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Theorem

Let $\left \langle{E_n}\right \rangle$ be the sequence of real functions $E_n: \R \to \R$ defined as:

$E_n \left({x}\right) = \left({1 + \dfrac x n}\right)^n$


Then, for each $x \in \R$ and for sufficiently large $n \in \N$, $E_n \left({x}\right)$ is positive.

That is:

$\forall x \in \R: \forall n \in \N: n \ge \left \lceil{\left \vert{x}\right \vert}\right \rceil \implies E_n \left({x}\right) > 0$

where $\left \lceil{x}\right \rceil$ denotes the ceiling of $x$.


Proof

Fix $x \in \R$.


Then:

\(\displaystyle n\) \(\ge\) \(\displaystyle \left \lceil{\left \vert{x}\right \vert}\right \rceil\)
\(\displaystyle \implies \ \ \) \(\displaystyle n\) \(>\) \(\displaystyle -x\) Real Number is between Ceiling Functions and Negative of Absolute Value
\(\displaystyle \implies \ \ \) \(\displaystyle 1\) \(>\) \(\displaystyle \frac{-x} n\) dividing both sides by $n$
\(\displaystyle \implies \ \ \) \(\displaystyle 1 + \frac x n\) \(>\) \(\displaystyle 0\) adding $\dfrac{-x} n$ to both sides
\(\displaystyle \implies \ \ \) \(\displaystyle \left({1 + \frac x n}\right)^n\) \(>\) \(\displaystyle 0\) Power of Strictly Positive Real Number is Strictly Positive: Positive Integer

$\blacksquare$