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:
Where is my logic failing? Has the inequality sense when the limit is applied?
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 $$\frac{1}{n} > 0 \implies \lim_{n \to \infty} \frac{1}{n} > \lim_{n \to \infty} 0 \implies 0 > 0.$$