ProofWiki:Sandbox

Theorem
Let $\left \langle{E_n}\right \rangle$ be the sequence of mappings $E_n: \R \to \R$ defined by $E_n(x) = \left({1 + \dfrac{x}{n}}\right)^n$.

Then, for sufficiently large $n \in \N$, $\left \langle{E_n(x)}\right \rangle$$ is increasing with respect to $n$.

That is:
 * $\forall x \in \R \setminus \left \{ {0} \right \} \exists N \in \N : n \geq N \implies E_n \left({ x }\right) < E_{n+1} \left({ x }\right)$

Proof
Fix some $x_0 \in \R \setminus \left \{ {0} \right \}$.

Let $N = \left \lceil{ \left|{ x_0 }\right| }\right \rceil + 1$, where $\left \lceil {\cdot} \right \rceil$ denotes the ceiling function.

Then, for $n \geq N$:
 * $1 + \dfrac{x_0}{n} > 0$

So we may apply the AM-GM inequality, with $x_1 = 1$ and $x_2 = \ldots = x_{n+1} = 1 + \dfrac{x_0}{n}$ to obtain that:
 * $\dfrac{ 1 + n \left({ 1 + \dfrac{x_0}{n} }\right) }{n + 1} > \left({ \left({ 1 + \dfrac{x_0}{n} }\right) ^n }\right) ^{ 1/ \left({ n+1 }\right) }$

Or, after some simplification:
 * $1 + \dfrac{x_0}{n+1} > \left({ 1 + \dfrac{x_0}{n} }\right)^{ \dfrac{n}{n+1} }$

From which it follows that
 * $ E_{n+1} \left({ x }\right) = \left({ 1 + \dfrac{x_0}{n+1} }\right) ^{n+1} > \left({ \left({ 1 + \dfrac{x_0}{n} }\right)^{ \dfrac{n}{n+1} } }\right)^{n+1} = \left({ 1 + \dfrac{x_0}{n} }\right) ^n = E_n \left({ x }\right)$

Hence the result.