# A contradiction or a wrong calculation? $3\lt\lim_{n\to\infty}\log_n(p)\lt3$, $\forall n$ (sufficiently large) where $n^3\lt p\in\Bbb P\lt(n+1)^3$?

This is probably a very trivial observation but at the same time somehow interesting to me. For every sufficiently large $$n$$, it has been already proved that:

$$\exists p \in \Bbb P: n^3 \lt p \lt (n+1)^3$$

Then applying $$\log_n$$ two all the terms:

$$\log_n n^3 \lt \log_np \lt \log_n(n+1)^3$$

By the properties of logarithms:

$$3 \lt \log_np \lt 3 \cdot \log_n(n+1)$$

But basically if the following limit is not wrong $$\lim_{n \to \infty}\log_n(n+1)=1$$

It would mean that when $$\lim_{n \to \infty}$$:

$$3 \lt \log_np \lt 3$$

But is not that expression a contradiction?

Instead, I was expecting something like this:

$$\lim_{n \to \infty} \log_np=3$$

For instance the same exercise using Bertrand's postulate does not have that problem:

$$\forall n \exists p: n \lt p \in \Bbb P \lt 2n$$

$$\land$$

$$1 \lt log_np \lt 2$$

So at least there exists a prime $$p$$ in a $$[n,2n]$$ interval such that:

$$\lim_{n \to \infty} \log_np \lt 2$$

I would like to ask the following questions:

1. Where is my logic failing? Has the inequality sense when the limit is applied?

2. Is there some literature regarding the study of this kind of limits? Thank you!

You cannot take the limit of an inequality. Consider the contradiction $$\frac{1}{n} > 0 \implies \lim_{n \to \infty} \frac{1}{n} > \lim_{n \to \infty} 0 \implies 0 > 0.$$
• so basically the point is that when we work with inequalities (in the way shown in the question and your answer) we just can say something about what happens when the value of $n$ gets very big, but not about the limit when it goes to infinity – iadvd Sep 23 '16 at 1:32
• I would instead say that you can take limits of inequalities (assuming the limits of both sides exist), but that strict inequalities can become non-strict inequalities in the process. In other words, if $\lim_{n\rightarrow \infty} a_n = a$ and $\lim_{n\rightarrow \infty } b_n = b$, then the inequality $a_n < b_n$ for all sufficiently large $n$ implies $a\leq b$. (But +1 from me anyway) – Jason DeVito Sep 23 '16 at 2:06