Derivative of Exponential Function/Proof 5/Lemma

Theorem

For $x \in\ R$, let $\left\lceil{x}\right\rceil$ denote the ceiling of $x$.

Then:

$\forall x \in \R : n \ge \left\lceil{\left\vert{x}\right\vert}\right\rceil \implies \left\langle{\dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle$ is increasing.

Proof

First:

 $\displaystyle n$ $>$ $\displaystyle \left\lceil{\left\vert{x}\right\vert}\right\rceil$ $\displaystyle \implies \ \$ $\displaystyle n$ $>$ $\displaystyle -x$ Negative of Absolute Value and Real Number is between Ceiling Functions $\displaystyle \implies \ \$ $\displaystyle \frac n {n + x}$ $>$ $\displaystyle 0$

Then:

 $\displaystyle n$ $>$ $\, \displaystyle \left\lceil{\left\vert{x}\right\vert}\right\rceil \,$ $\displaystyle$ $\displaystyle \implies \ \$ $\displaystyle n$ $>$ $\, \displaystyle \left\vert{x}\right\vert \,$ $\displaystyle$ Real Number is between Ceiling Functions $\displaystyle \implies \ \$ $\displaystyle \left\vert{\frac x n}\right\vert$ $<$ $\, \displaystyle 1 \,$ $\displaystyle$ dividing both sides by $n$ $\displaystyle \implies \ \$ $\displaystyle -1$ $<$ $\, \displaystyle \frac x n \,$ $\, \displaystyle <\,$ $\displaystyle 1$ Negative of Absolute Value: Corollary 1 $\displaystyle \implies \ \$ $\displaystyle 0$ $<$ $\, \displaystyle 1 + \frac x n \,$ $\displaystyle$ $\displaystyle \implies \ \$ $\displaystyle 0$ $<$ $\, \displaystyle \left({1 + \frac x n}\right)^n \,$ $\displaystyle$ Power of Positive Real Number is Positive: Natural Number

So, for $n \ge \left\lceil{\left\vert{x}\right\vert}\right\rceil$:

$\left\langle{\dfrac n {n + x} }\right\rangle$

and:

$\left\langle{\left({1 + \dfrac x n}\right)^n}\right\rangle$

are positive.

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

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

Then:

 $\displaystyle \frac n {n + x}$ $\le$ $\displaystyle \frac {n + 1} {n + x + 1}$ $\displaystyle \iff \ \$ $\displaystyle n \left({n + x + 1}\right)$ $\le$ $\displaystyle \left({n + 1}\right) \left({n + x}\right)$ Real Number Ordering is Compatible with Multiplication $\displaystyle \iff \ \$ $\displaystyle n^2 + n x + n$ $\le$ $\displaystyle n^2 + n x + n + x$ $\displaystyle \iff \ \$ $\displaystyle 0$ $\le$ $\displaystyle x$

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.
$n \ge \left\lceil{\left\vert{x}\right\vert }\right\rceil \implies \left\langle{\dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle$ is increasing.

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

Aiming for a contradiction, suppose that:

$\left\langle{\dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle$ is decreasing.

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

$\left\langle{\dfrac {n + x} n \dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle = \left\langle{\left({1 + \dfrac x n}\right)^n}\right\rangle$ is decreasing.

Hence the result, by Proof by Contradiction.

$\blacksquare$