Exponential Sequence is Eventually Increasing
Jump to navigation
Jump to search
Theorem
Let $\sequence {E_n}$ be the sequence of real functions $E_n: \R \to \R$ defined as:
- $\map {E_n} x = \paren {1 + \dfrac x n}^n$
Then, for sufficiently large $n \in \N$, $\sequence {\map {E_n} x}$ is increasing with respect to $n$.
That is:
- $\forall x \in \R: \forall n \in \N: n \ge \ceiling {\size x} \implies \map {E_n} x \le \map {E_{n + 1} } x$
where $\ceiling x$ denotes the ceiling of $x$.
Proof
Fix $x \in \R$.
Then:
\(\ds n\) | \(\ge\) | \(\ds \ceiling {\size x}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds n\) | \(>\) | \(\ds -x\) | Real Number is between Ceiling Functions and Negative of Absolute Value | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1\) | \(>\) | \(\ds \frac {-x} n\) | Divide both sides by $n$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1 + \frac x n\) | \(>\) | \(\ds 0\) |
So we may apply the AM-GM inequality, with $x_1 := 1$ and $x_2 := \ldots := x_{n + 1} = 1 + \dfrac x n$, to obtain that:
This article, or a section of it, needs explaining. In particular: The above sentence is unclear. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
- $\dfrac {1 + n \paren {1 + \dfrac x n} } {n + 1} > \paren {\paren {1 + \dfrac x n}^n}^{1 / \paren {n + 1} }$
After simplification:
- $1 + \dfrac x {n + 1} > \paren {1 + \dfrac x n}^{n / \paren {n + 1} }$
From Power Function is Strictly Increasing over Positive Reals: Natural Exponent:
- $\paren {1 + \dfrac x {n + 1} }^{n + 1} > \paren {1 + \dfrac x n}^n$
where we have raised both sides to the power of $n + 1$.
Hence the result.
$\blacksquare$