Curious case of asymptotic equivalence

Question

Let us consider $n$ a positive integer and $F(n)$ an increasing function of $n$.

What conditions $F$ should fulfil to obtain the following:

$$\lim_{n \to \infty} \frac{F(n)}{F(n-1)} = 1$$

?

It appears that this is wrong if $F(n)=exp(n)$.

Are there general results about this ?

Thank you.

@5xum Thank you for your comment. Are we able to say something for other types of function? — Dingo13, Jun 30 '17 at 12:14
If $g(n)$ satisfies it, then $g(n)P(n)$ satisfies it, where $P(n)$ is a polynomial. — Simply Beautiful Art, Jun 30 '17 at 12:56
You also have the trivial case of $F$ bounded, i.e. $\lim_{n\to\infty}F(n)$ exists. — Miguel, Jun 30 '17 at 13:10

score 1 · Answer 1 · answered Jun 30 '17 at 13:02

1

Just a comment as a contribution, is that I believe that you could be interested in Theorem 8, from Jakimczuk, Functions of Slow Increase and Integer Sequences, Journal of Integer Sequences, Vol. 13 (2010), Article 10.1.1. It is an open access journal.

answered Jun 30 '17 at 13:02

Thank you for your link. – Dingo13 Jun 30 '17 at 16:26
@Dingo13 I know that this journal is very good. Enjoy it! – Jun 30 '17 at 16:28

Curious case of asymptotic equivalence

1 Answers1