I understand that for two functions $f(n)$ and $g(n)$ for $n\in\mathbb N$, the notation $$g(n)=o(f(n)) \mbox{ as } n\rightarrow \infty$$ means that for all $\epsilon>0$, there exists $N\in\mathbb N$ such that $|g(n)|\le \epsilon |f(n)|$ whenever $n>N$, or equivalently, if $f(n)$ is non-zero for sufficiently large $n$ $$\lim_{n\rightarrow \infty} \dfrac{g(n)}{f(n)}=0.$$ And variants of the notation for real functions exist too.

However, I have seen many people who treats "$o(n)$" as some quantity. For example, they write things like $$g(n)=\left(f(n)+o\left(1/n\right)\right)^n.$$ This doesn't seem to make sense at all just looking at the definition. Can someone give me a rigorous definition of what an expression of this sort really means?

  • $\begingroup$ In those case, then mean "for some $h$ with $h(n)=f(n)+o(1/n)$, $g(n)=h(n)^n$." Or, equivalently, that $g(n)^{1/n}-f(n)=o(1/n)$. It is definitely abuse of the original notation (and the original notation abuses the "=" sign.) $\endgroup$ Nov 22, 2015 at 21:53
  • $\begingroup$ @ThomasAndrews why does it abuse the original notation? One could say the entire point of introducing that type of notation and writing it with = is to write things as asked about. $\endgroup$
    – quid
    Nov 22, 2015 at 21:58

2 Answers 2


This means that $ g(n) = (f(n) + h(n))^n $ for some function $ h$ such that $h(n) = o(1/n) $.

  • 1
    $\begingroup$ I see. I'm surprised that I couldn't find any explanation of this on the internet despite its common usage. Thanks! (I love your name btw) $\endgroup$
    – Gawin
    Nov 22, 2015 at 22:09

This expression means that $g(n)=\left(f(n)+h(n)\right)^n$ where $h(n)$ is some (unspecified) function that is $o(1/n)$.


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