Exponential Sequence is Eventually Strictly Positive

From ProofWiki
Jump to navigation Jump to search

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

$\blacksquare$