Equivalence of Definitions of Asymptotically Equal Sequences

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.

$\blacksquare$