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:

Then:

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:

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 $x \in \R_{<0}$.


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