Does this limit imply that a function is "close" to Lambert W?

Question

Suppose I am given the following limit involving function $f(n)\geq 0$:

$$\lim_{n\rightarrow\infty}\log n-f(n)-\log f(n)=c$$

where $c$ is a constant.

I am wondering if that implies that $f(n)$ is "close to" Lambert W function $W(ne^{-c})$ in the following sense:

$$\lim_{n\rightarrow\infty}|f(n)-W(ne^{-c})|=0$$

I think that it is true by the following logic, however, I am not sure about the last step (or whether the logic is correct in the first place):

$$\begin{array}{ll} &\lim_{n\rightarrow\infty}\log n-f(n)-\log f(n)=c\\ \Rightarrow&\forall\epsilon>0~\exists n_0\text{ such that } \forall n\geq n_0:\\ &c-\epsilon\leq\log n-f(n)-\log f(n)\leq c+\epsilon \text{ by the definition of the limit}\\ \Rightarrow&e^{c-\epsilon}\leq\frac{n}{f(n)e^{f(n)}}\leq e^{c+\epsilon}\\ \Rightarrow&ne^{-c-\epsilon}\leq f(n)e^{f(n)}\leq ne^{-c+\epsilon}\\ \Rightarrow&W(ne^{-c-\epsilon})\leq f(n)\leq W(ne^{-c+\epsilon}) \text{ by the fact that }f(n)\geq 0 \end{array}$$

Up to this point, I am sure of my steps. Here is where it gets complicated for me. I perform the Taylor series expansion around $\epsilon=0$ to obtain the following:

$$W(ne^{-c-\epsilon})=W(ne^{-c})-\epsilon\frac{W(ne^{-c})}{1+W(ne^{-c})}+\epsilon^2\frac{W(ne^{-c})}{2(1+W(ne^{-c}))^3}+\mathcal{O}(\epsilon^3)$$ $$W(ne^{-c+\epsilon})=W(ne^{-c})+\epsilon\frac{W(ne^{-c})}{1+W(ne^{-c})}+\epsilon^2\frac{W(ne^{-c})}{2(1+W(ne^{-c}))^3}+\mathcal{O}(\epsilon^3)$$

Now, the first term in the expansion is what I need to subtract from the entire inequality and the second term in the expansion can be upper-bounded by $\epsilon$ since $\frac{W(x)}{1+W(x)}\leq 1$ when $x\geq 0$. If I could somehow justify ignoring the third term, then I'd be done, as that would imply: $$-\epsilon\leq f(n)-W(ne^{-c})\leq \epsilon$$ However, I am not sure if I can do that, or, if I can, how to justify that step.

Can anyone help?

Answer 1

You know, I don't think you need the Taylor's series. You already have $W(ne^{-c-\epsilon}) \le f(n) \le W(ne^{-c+\epsilon})$. Subtract $W(ne^{-c})$ all the way through your inequality. You get

$W(ne^{-c-\epsilon}) - W(ne^{-c}) \le f(n) - W(ne^{-c}) \le W(ne^{-c+\epsilon}) - W(ne^{-c})$.

The left and right sides of this inequality both go to 0 as $\epsilon \rightarrow 0$. What more do you need?

I think you really had it solved -- I've put this in the answer section because it was kind of hard to write in a comment.

Answer 2

Your steps are valid, and as Betty mentioned you want to look at

$$ W(ne^{-c-\epsilon}) - W(ne^{-c}) \le f(n) - W(ne^{-c}) \le W(ne^{-c+\epsilon}) - W(ne^{-c}), $$

which holds for all $n \geq n_1$, say.

It is a consequence of the asymptotic formula for the Lambert function that

$$ \lim_{x \to \infty} W(\alpha x) - W(x) = \log \alpha, $$

And for this problem we have $x = ne^{-c}$ and $\alpha = e^{\pm \epsilon}$. This means that we can find an $n_2$ such that

$$ W(ne^{-c+\epsilon}) - W(ne^{-c}) \le \epsilon + \log e^\epsilon = 2\epsilon $$

and

$$ -2\epsilon = -\epsilon + \log e^{-\epsilon} \leq W(ne^{-c-\epsilon}) - W(ne^{-c}) $$

for all $n \geq n_2$. Then if we take $n \geq \max\{n_1,n_2\}$ we have

$$ -2\epsilon \leq f(n) - W(ne^{-c}) \leq 2\epsilon, $$

so that

$$ \lim_{n \to \infty} f(n) - W(ne^{-c}) = 0. $$