Exponential Sequence is Eventually Increasing

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 $x$.

Fix $x \in \R$.

Then:

$\ds n$

$\ge$

$\ds \ceiling {\size x}$

$\ds \leadsto \ \ $

$\ds n$

$>$

$\ds -x$

$\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:

$x_1 := 1$

$x_j := 1 + \dfrac x n$ for each $j \in \set {2, 3, \dotsc, n + 1}$

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:

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

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

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

Hence the result.

$\blacksquare$