Exponential Sequence is Eventually Strictly Positive
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Theorem
Let $\sequence {E_n}$ be the sequence of real functions $E_n: \R \to \R$ defined as:
- $\map {E_n} x = \paren {1 + \dfrac x n}^n$
Then, for each $x \in \R$ and for sufficiently large $n \in \N$, $\map {E_n} x$ is positive.
That is:
- $\forall x \in \R: \forall n \in \N: n \ge \ceiling {\size x} \implies \map {E_n} x > 0$
where $\ceiling x$ denotes the ceiling of $x$.
Proof
Fix $x \in \R$.
Then:
\(\ds n\) | \(\ge\) | \(\ds \ceiling {\size x}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds n\) | \(>\) | \(\ds -x\) | Real Number is between Ceiling Functions and Negative of Absolute Value | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1\) | \(>\) | \(\ds \frac {-x} n\) | dividing both sides by $n$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1 + \frac x n\) | \(>\) | \(\ds 0\) | adding $\dfrac {-x} n$ to both sides | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {1 + \frac x n}^n\) | \(>\) | \(\ds 0\) | Power of Strictly Positive Real Number is Strictly Positive: Positive Integer |
$\blacksquare$