Suppose that $f : \mathbb{N} \to \mathbb{R}_{++}$, $g : \mathbb{N} \to \mathbb{R}_{++}$ are such that $F := \lim_{n \to \infty} f(n)$ and $G := \lim_{n \to \infty} g(x)$ both exist. Let $0<a,b<1$ and let $\lceil \cdot \rceil$ denote the ceiling operator.

Claim. Suppose that

$$\lim_{n \to \infty} \frac{(f(n))^{\lceil an \rceil}}{(g(n))^{\lceil b n \rceil}} = c $$

where $0<c<\infty$.

Then the following hold:

  1. By the density of $\mathbb{Q}$ in $\mathbb{R}$, there exist $k,l<n$ such that $\lim_{n \to \infty} k/n = a$ and $\lim_{n \to \infty} l/n = b$. Therefore

$$ \lim_{n \to \infty} \frac{(f(n))^{\lceil an \rceil}}{(g(n))^{\lceil b n \rceil}} = \lim_{n \to \infty} \left( \frac{(f(n))^a}{(g(n))^b} \right)^n$$

  1. $F^a = G^b$ Why? Otherwise, the limit above would be either $0$ (when $F^a<G^b$) or positive infinity (when $F^a>G^b$).

All of these seem intuitively true to me. Is there anything I should be careful about or can I safely assume these two facts going forward?


1 Answer 1


I would reason like this: $\frac{[an]}{[bn]}=\frac{an-O(1)}{bn-O(1)} \to a/b$ so $\frac{[an]}{[bn]}\log f(n)-\log g(n) \to a\log F/b-\log G$; but $\log c/[bn]\to 0$ so the hypothesis implies (taking log) that $\frac{[an]}{[bn]}\log f(n)-\log(n) \to 0$ so $a\log F=b\log G$

Note that any $a,b >0$ will work too


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