Equivalence of Definitions of Asymptotically Equal Sequences
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Theorem
Let $\sequence {a_n}$ and $\sequence {b_n}$ be sequences in $\R$.
The following definitions of the concept of Asymptotically Equal Sequences are equivalent:
Definition 1
Let $b_n \ne 0$ for all $n$.
$\sequence {a_n}$ is asymptotically equal to $\sequence {b_n}$ if and only if:
- $\ds \lim_{n \mathop \to \infty} \dfrac {a_n} {b_n} = 1$
Definition 2
$\sequence {a_n}$ is asymptotically equal to $\sequence {b_n}$ if and only if:
- $a_n - b_n = \map o {b_n}$
where $o$ denotes little-$\oo$ notation.
Definition 3
$\sequence {a_n}$ is asymptotically equal to $\sequence {b_n}$ if and only if:
- $a_n - b_n = \map \oo {a_n}$
where $\oo$ denotes little-$\oo$ notation.
Proof
$(1)$ iff $(2)$
\(\ds \lim_{n \mathop \to \infty} \dfrac {a_n} {b_n}\) | \(\to\) | \(\ds 1\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \lim_{n \mathop \to \infty} \paren {\dfrac {a_n} {b_n} - \dfrac {b_n} {b_n} }\) | \(\to\) | \(\ds 0\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \lim_{n \mathop \to \infty} \dfrac {a_n - b_n} {b_n}\) | \(\to\) | \(\ds 0\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds a_n - b_n\) | \(=\) | \(\ds \map \oo {b_n}\) |
$\Box$
Definition $(2)$ iff $(3)$
Let $a_n - b_n = \map \oo {b_n}$.
Let $0 < \epsilon < 1/2$.
Then:
\(\ds \epsilon \cdot \size {b_n}\) | \(\ge\) | \(\ds \size {a_n - b_n}\) | For $n$ sufficiently large | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {1 - \epsilon} \size {a_n - b_n} + \epsilon \cdot \size {a_n - b_n}\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds \paren {1 - \epsilon} \size {a_n - b_n} - \epsilon \cdot \size {a_n} + \epsilon \cdot \size {b_n}\) | Triangle Inequality |
So:
- $\size {a_n - b_n} \le \dfrac {\epsilon \cdot \size {a_n} } {1 - \epsilon} \le 2 \epsilon \cdot \size {a_n}$
Thus:
- $a_n - b_n = \map \oo {a_n}$
The other implication follows by symmetry.
$\blacksquare$