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

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

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