Derivative of Exponential Function/Proof 5/Lemma

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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:

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


Then:

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

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


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.


Hence, from Product of Positive Increasing Functions 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.


Thus, from Product of Positive Increasing and Decreasing Functions 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.

This contradicts Exponential Sequence is Eventually Increasing.


Hence the result, by Proof by Contradiction.

$\blacksquare$