Derivative of Exponential Function/Proof 5/Lemma

Theorem
Let $\left\lceil{ \cdot }\right\rceil$ denote the ceiling function.

Then:
 * $\displaystyle \forall x \in \R : n \geq \left\lceil{ \left\vert{ x }\right\vert }\right\rceil \implies \left\langle{ \frac{ n }{ n + x } \left({ 1 + \frac{x}{n} }\right)^{n} }\right\rangle$ is increasing

Proof
First:

And:

So, for $n \geq \left\lceil{ \left\vert{ x }\right\vert }\right\rceil$, $\left\langle{ \dfrac{ n }{ n + x } }\right\rangle$ and $\left\langle{ \left({ 1 + \dfrac{x}{n} }\right)^{n} }\right\rangle$ are  positive.

Now let $n \geq \left\lceil{ \left\vert{ x }\right\vert }\right\rceil$.

Suppose first that $x \in \R_{\geq 0}$.

Then:

So $\left\langle{ \dfrac{ n }{ n + x } }\right\rangle$ is increasing.

Further, from Exponential Sequence is Eventually Increasing: $\left\langle{ \left({ 1 + \dfrac{x}{n} }\right)^{n} }\right\rangle$ is increasing.

Hence, from Product of Positive Increasing Functions is Increasing:
 * $\displaystyle n \geq \left\lceil{ \left\vert{ x }\right\vert }\right\rceil \implies \left\langle{ \frac{ n }{ n + x } \left({ 1 + \frac{x}{n} }\right)^{n} }\right\rangle$ is increasing

Suppose instead that $x \in \R_{<0}$.

that:
 * $\displaystyle \left\langle{ \frac{ n }{ n + x } \left({ 1 + \frac{x}{n} }\right)^{n} }\right\rangle$ is decreasing

From above: $\left\langle{ 1 + \dfrac{ x }{ n } }\right\rangle = \left\langle{ \dfrac{ n + x }{ n } }\right\rangle$ is decreasing.

Thus, from Product of Positive Increasing and Decreasing Functions is Decreasing:
 * $\displaystyle \left\langle{ \frac{ n + x }{ n } \frac{ n }{ n + x } \left({ 1 + \frac{x}{n} }\right)^{n} }\right\rangle = \left\langle{ \left({ 1 + \frac{x}{n} }\right)^{n} }\right\rangle$ is decreasing.

This contradicts Exponential Sequence is Eventually Increasing.

Hence the result, by Proof by Contradiction.