Does there exist a function $f(n)$ that will equal to $n \cdot\Theta(f(n))$?

Question

Does there exist a function $f(n)$ that will equal to $n \cdot\Theta(f(n))$?

I thought about taking $\cfrac{1}{n^2}$ and then both of them will be (for high values of $n$) $\rightarrow 0$

What do you guys think?

Answer 1

No: $anf(n) \le f(n) \le bnf(n)$ with $f(n)>0$ and $0<a<b$ implies $an\le 1 \le bn$ for all $n$.