Derivative of Exponential Function/Proof 5/Lemma
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.
$\blacksquare$