# Does the limit exist and equal $1$?

Let $$n\ge m>1$$ be integers. I want to know whether the following limit exists:

$$\lim\limits_{k\to \infty}\frac{q_1q_2\cdots q_{[k\log_n m]}}{m^{\log_n q_1q_2\cdots q_k}},$$ where $$q_i (i\ge 1)$$ are common factors of $$m$$ and $$n$$ (i.e., $$q_i | n$$ and $$q_i |m$$ for all $$i\ge 1$$) and $$[x]$$ denotes the integer part of $$x$$.

I guess the limit is $$1$$ and can show that by some special examples (for example, all $$q_i$$ are the same , or $$n=m$$ and so on), but I cannot prove it. Can someone help me? Many thanks!

• So the $q_i$ is just some arbitrary sequence of divisors of $(m,n)$? – Erick Wong Sep 3 '20 at 2:21
• Yes, $q_i$ are chosen from the common factors of $m$ and $n$ and they are maybe equal at every step. – ljjpfx Sep 3 '20 at 2:23
• It’s probably cleaner to write the denominator as $(q_1\cdots q_k)^c$, where $c := \log_n m \le 1$. It would be instructive to look at a simple case like $c=1/2$. I think you can construct sequences of $q_i$ where it wobbles far above and below $1$ and fails to converge. – Erick Wong Sep 3 '20 at 2:43
• Thank you @Erick Wong. Then, can we show that the limit inferior is greater than $0$? – ljjpfx Sep 3 '20 at 3:05
• I believe that lim sup $=\infty$ and lim inf $=0$ can be achieved for the same sequence. Shall I write an answer? – Erick Wong Sep 3 '20 at 3:07

## 1 Answer

I believe the answer is no (for some choice of $$g_i$$) except when $$(m,n)=1$$ or $$m=n$$. It is helpful to define $$c := \log_n m \le 1$$. Then the expression in the limit can be written as:

$$Q_k = \frac{q_1 \cdots q_{\lfloor ck\rfloor}}{(q_1 \cdots q_k)^c}$$

We claim that, if $$0 < c<1$$ and $$q_i$$ can take at least one other value $$a>1$$, then there exists a choice of sequence $$q_i$$ that makes $$\liminf_k Q_k = 0$$ and $$\limsup_k Q_k = \infty$$.

First, note that we can make a single value of $$Q_k$$ as small or as large as we like (by choosing a suitable $$k$$). As $$k\to \infty$$ there are an unbounded number of factors in the denominator which are not present at all in the numerator. By setting all the $$q_i = 1$$ up to $$i \le \lfloor ck \rfloor$$, and all remaining $$q_i = a$$, we can get an unbounded number of $$a$$s on the denominator, raised to a fixed positive power, making the fraction as close to $$0$$ as we like.

Oppositely, by stuffing $$q_i = a$$ up to $$i \le \lfloor ck \rfloor$$ and $$q_i = 1$$ everywhere else, we get an unbounded number of $$a$$s raised to the power $$1-c$$, which can be as large as we like.

At this point you should be able to convince yourself that we can still achieve these lows and highs, given any finite prefix of prescribed $$q_1, q_2, \ldots, q_s$$. We just pick our $$k$$ large enough (relative to $$s$$) so that the contribution from our well-chosen $$a$$ and $$1$$ terms outweighs the contribution from the first $$s$$ terms. It is instructive to consider a simple model like the case $$c=1/2$$ where the exponents and floor function are cleaner. The details are a bit messy, I don’t really wish to work them out unless necessary.

Once we have this, we can obtain a single sequence of $$q_i$$ as follows: pick some initial prefix giving $$Q_k = 1/2$$. Then extend it out so that $$Q_{k’} = 2$$. Then extend it again so that $$Q_{k’’}= 1/3$$, then $$3$$, $$1/4$$, $$4$$, and so on. This will take the shape of increasingly (exponentially) longer stretches of $$1$$s and $$a$$s.

Morally, these are exponentially large values we are multiplying and dividing, and there is a significant lag between the numerator and denominator (you can change a lot of terms in the denominator without affecting the numerator until much later). So it’s not surprising that it doesn’t converge unconditionally.