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:

\(\displaystyle n\) \(>\) \(\displaystyle \left\lceil{\left\vert{x}\right\vert}\right\rceil\)
\(\displaystyle \implies \ \ \) \(\displaystyle n\) \(>\) \(\displaystyle -x\) 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.


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.

$\blacksquare$