TITLE: Name for kind of big O notation with leading coefficient
QUESTION [6 upvotes]: Context: As known the big O notation $O(f(n))$ describes a function $g(n)$ such that there is a constant $C \ge 0$ with $\limsup_{n\to\infty} \left|\frac{g(n)}{f(n)}\right| \le C$ (I assume that $f(n)$ is never zero). Thus the big O notation can be used to characterize the speed of convergence (for example for an algorithm). So I can write $$a(n) = a + O\left(\frac 1n\right)$$ which means that the function / algorithm $a(n)$ calculates in almost all calculation steps $n$ the desired number $a$ with an error less than $\frac Cn$ for a constant $C\ge 0$.
Problem: Whereby I can compare two algorithms by their speed of convergence with the big O notation, the above notation $a(n) = a + O\left(\frac 1n\right)$ does not say anything about the actual error in the n-th step because I do not know the constant $C\ge 0$.
My solution: One may introduce a new notation, lets say the big Psi notation $\Psi(f(n))$ which is defined as
$$g(n)\in \Psi(f(n)) \iff \limsup_{n\to\infty} \left|\frac{g(n)}{f(n)}\right| < 1$$
For example one may write $$a(n) = a + \Psi\left(\frac{42}{n}\right)$$ so that $a(n)\in O\left(\frac 1n\right)$ with the constant $C=42$. There are also arithmetic rules for the big Psi notation similar to the rules for the big notation, for example:
$$\left(1+\Psi\left(\frac an\right)\right)\left(1+\Psi\left(\frac bn\right)\right)\subseteq 1+\Psi\left(\frac{a+b+ab}{n}\right)$$
My question: Because the proposed solution is simple and somehow straightforward I guess there was already a mathematician who wrote about it. Can you point me to a textbook/paper where this notation is discussed, please? How is this notation called in mathematics?
So far I have only found the big Omega notation and the big Theta notation...
Update: After rethinking the notation I would now define
$$g(n)\in \Psi(f(n)) \iff \limsup_{n\to\infty} \left|\frac{g(n)}{f(n)}\right| \le 1$$
such that
$$\left(1+\Psi\left(\frac an\right)\right)\left(1+\Psi\left(\frac bn\right)\right)\subseteq 1+\Psi\left(\frac{a+b}{n}\right)$$
because
$$\limsup_{n\to\infty} \frac{\frac{a+b}{n}}{\frac{a+b}{n}+\frac{ab}{n^2}} = 1$$

REPLY [2 votes]: I don't think that
your $\Psi$ notation
has any advantage over
$a(n) 
\le a+\frac1{n}
$.
It does not really hide anything,
which the "O" and "o" notations
excel at.
Also, your equation
$$\left(1+\Psi\left(\frac an\right)\right)\left(1+\Psi\left(\frac bn\right)\right)\subseteq 1+\Psi\left(\frac{a+b+ab}{n}\right)
$$
is not correct,
or at least more imprecise than needed,
because
$(1+\frac{a}{n})(1+\frac{b}{n})
=1+\frac{a+b}{n}+\frac{ab}{n^2}
$.
The problem is
in the $ab$ term.
If it goes over $n$,
then the term is 
larger than necessary,
and if it is put over
$n^2$,
that term will get swallowed
by
$\frac{a+b+\epsilon}{n}$
for any $\epsilon > 0$.