Derivative of Exponential Function/Proof 5/Lemma

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Theorem

For $x \in\ R$, let $\ceiling x$ denote the ceiling of $x$.


Then:

$\forall x \in \R : n \ge \ceiling {\size x} \implies \sequence {\dfrac n {n + x} \paren {1 + \dfrac x n}^n}$ is increasing.


Proof

First:

\(\ds n\) \(>\) \(\ds \ceiling {\size x}\)
\(\ds \leadsto \ \ \) \(\ds n\) \(>\) \(\ds -x\) Negative of Absolute Value and Real Number is between Ceiling Functions
\(\ds \leadsto \ \ \) \(\ds \frac n {n + x}\) \(>\) \(\ds 0\)


Then:

\(\ds n\) \(>\) \(\, \ds \ceiling {\size x} \, \) \(\ds \)
\(\ds \leadsto \ \ \) \(\ds n\) \(>\) \(\, \ds \size x \, \) \(\ds \) Real Number is between Ceiling Functions
\(\ds \leadsto \ \ \) \(\ds \size {\frac x n}\) \(<\) \(\, \ds 1 \, \) \(\ds \) dividing both sides by $n$
\(\ds \leadsto \ \ \) \(\ds -1\) \(<\) \(\, \ds \frac x n \, \) \(\, \ds < \, \) \(\ds 1\) Negative of Absolute Value: Corollary 1
\(\ds \leadsto \ \ \) \(\ds 0\) \(<\) \(\, \ds 1 + \frac x n \, \) \(\ds \)
\(\ds \leadsto \ \ \) \(\ds 0\) \(<\) \(\, \ds \paren {1 + \frac x n}^n \, \) \(\ds \) Power of Positive Real Number is Positive: Natural Number


So, for $n \ge \ceiling {\size x}$:

$\sequence {\dfrac n {n + x} }$

and:

$\sequence {\paren {1 + \dfrac x n}^n}$

are positive.


Now let $n \ge \ceiling {\size x}$.


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


Then:

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


So $\sequence {\dfrac n {n + x} }$ is increasing.

Further, from Exponential Sequence is Eventually Increasing:

$\sequence {\paren {1 + \dfrac x n}^n}$ is increasing.


Hence, from Product of Positive Increasing Functions is Increasing:

$n \ge \ceiling {size x} \implies \sequence {\dfrac n {n + x} \paren {1 + \dfrac x n}^n}$ is increasing.


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


Aiming for a contradiction, suppose:

$\sequence {\dfrac n {n + x} \sequence {1 + \dfrac x n}^n}$ is decreasing.

From above: $\sequence {1 + \dfrac x n} = \sequence {\dfrac {n + x} n}$ is decreasing.


Thus, from Product of Positive Increasing and Decreasing Functions is Decreasing:

$\sequence {\dfrac {n + x} n \dfrac n {n + x} \paren {1 + \dfrac x n}^n} = \sequence {\paren {1 + \dfrac x n}^n}$ is decreasing.

This contradicts Exponential Sequence is Eventually Increasing.


Hence the result, by Proof by Contradiction.

$\blacksquare$