Is $\sqrt{2/(27\pi n)}\sim n^{-1/2}$?

Question

Is $\sqrt{2/(27\pi n)}\sim n^{-1/2}$?

Since $$ \sqrt{\frac{2}{27\pi n}}=\sqrt{\frac{2}{27\pi}}\cdot\frac{1}{\sqrt{n}}\sim\frac{1}{\sqrt{n}}=n^{-1/2}, $$ I would say, yes, of course.

Answer 1

It depends on how you define the relation $\sim$. It is certainly the case that $\sqrt{2/(27\pi n)} = \Theta(n^{-1/2})$.

But if you define $f\sim g$ to mean that $f(n)/g(n)\to 1$ as $n\to\infty$, which is how $\sim$ is usually defined, then it is not true as the limit of the quotient of $\sqrt{2/(27\pi n)}$ and $n^{-1/2}$ in your case above is not $1$.

See Asymptotic Notation.