Exponential Sequence is Eventually Increasing

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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 sufficiently large $n \in \N$, $\sequence {\map {E_n} x}$ is increasing with respect to $n$.

That is:

$\forall x \in \R: \forall n \in \N: n \ge \ceiling {\size x} \implies \map {E_n} x \le \map {E_{n + 1} } x$

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
\(\ds \leadsto \ \ \) \(\ds 1\) \(>\) \(\ds \frac {-x} n\) Divide both sides by $n$
\(\ds \leadsto \ \ \) \(\ds 1 + \frac x n\) \(>\) \(\ds 0\)


So we may apply the AM-GM inequality, with $x_1 := 1$ and $x_2 := \ldots := x_{n + 1} = 1 + \dfrac x n$, to obtain that:



$\dfrac {1 + n \paren {1 + \dfrac x n} } {n + 1} > \paren {\paren {1 + \dfrac x n}^n}^{1 / \paren {n + 1} }$

After simplification:

$1 + \dfrac x {n + 1} > \paren {1 + \dfrac x n}^{n / \paren {n + 1} }$

From Power Function is Strictly Increasing over Positive Reals: Natural Exponent:

$\paren {1 + \dfrac x {n + 1} }^{n + 1} > \paren {1 + \dfrac x n}^n$

where we have raised both sides to the power of $n + 1$.

Hence the result.

$\blacksquare$