Limit of a sequence involving floor function

Question

I am reading an article, and in a certain point I need to estimate the following limit

$$\xi_p=\lim \limits_{k \to +\infty} \frac{k}{\lfloor\frac{k}{n}\rfloor+1},$$

where $n \ge 1$ is fixed.

The claim of the author is that $\xi_p < n$. I can have an intuition why that's true, but I can't formalize this. The best bound I can reach is $$\frac{k}{\lfloor\frac{k}{n}\rfloor+1} <\frac{k}{\frac{k-1}{n}+1}=\frac{kn}{k+n-1}$$

Therefore $\xi_p \le n$. What am I missing? Any help would be appreciated.

EDIT: I am adding some context. Maybe I am making a mistake in a point before that.

The article is about an upper bound for the percoletion correlation lenght on $\mathbb{Z}^d$. In a certain point, the authors reach to the bound $$P_p[0 \leftrightarrow \partial\Lambda_k] \le e^{-\lfloor k/n \rfloor-1}$$

where $\partial\Lambda_k$ is the boundary of the box centered on the origin with width $k$.

Just after that ist he claim that this implies that the correlation lenght $\xi_p$ is strictly less than $n$, and this is what I am trying to proof.

Since $$\xi_p= \lim \limits_{k \to +\infty} -\frac{k}{\log P_p[0 \leftrightarrow \partial\Lambda_k]}$$

and using the bound found earlier for the relevant probability, I arrived on the limit in the beginning of the question.

Here is the article in question. The relevant bound is on page 6. I guess I am makin a very dumb error, but i can't found it out.

Answer 1

Whether that is the floor function or the ceiling function, the limit is $n$, by a simple application of the sandwich theorem:

$$ k/n < \lfloor k/n \rfloor + 1 \le \lceil k/n \rceil + 1 \le k/n + 2 $$

So

$$ n = {k \over k/n} > {k \over \lfloor k/n \rfloor + 1} \ge {k \over \lceil k/n \rceil + 1} \ge {k \over k/n + 2} = {kn \over k + 2n}$$

But the limit of the last term

$$\lim_{k \to +\infty} {kn \over k+2n} = n$$

So all the limits are equal to $n$.

BTW it is true that for any finite $k$, the expression is $< n$ (whether that function is floor or ceiling). But that does not at all mean the limit $< n$.

Answer 2

This doesn't hold for the floor function.If it were the ceiling function, then,

$\lceil k/n\rceil \geq k/n$. So,$1+\lfloor k/n\rfloor \geq 1+ k/n> k/n$. This implies,

$n/k > \frac{1}{1+\lceil k/n\rceil}$ which gives $n> \frac{k}{1+\lceil k/n\rceil}$ (assuming $k$ is non-negative). Leting $k \rightarrow \infty$, you get the desired result for $n>1$.

Answer 3

If this is the floor function, then the claim is incorrect, as $$\frac{k}{\lfloor\frac{k}{n}\rfloor+1}\geq \frac{k}{\frac{k}{n}+1} = \frac{kn}{k+n} \to n \text{ as } k\to\infty$$