If $x$ and $y$ are asymptotically equal then $x(n)=y(n)+o(y(n))$.

Question

Show that

$x(n)\sim y(n)$ iff $x(n)=y(n)+o(y(n))$, where ' o ' here is "small o".

I used the following that I found from online source to prove it

$x(n)=y(n)(1+o(1))$, but I need to show it using definitions.

Can someone help me with this using definitions instead of my approach?

Answer 1

Suppose that $x(n)/y(n)$ is well defined for all sufficiently large values of $n$. Write $x(n) =y(n)+\varepsilon(n)$. Then $$ x(n) \sim y(n) \Leftrightarrow \frac{{x(n)}}{{y(n)}} \to 1 \Leftrightarrow \frac{{y(n) + \varepsilon (n)}}{{y(n)}} \to 1 \\ \Leftrightarrow 1 + \frac{{\varepsilon (n)}}{{y(n)}} \to 1 \Leftrightarrow \frac{{\varepsilon (n)}}{{y(n)}} \to 0 \Leftrightarrow \varepsilon (n) = o(y(n)). $$