Limit involving primes numbers

Question

For each integer $n$ denote by $p_{n}$ the largest prime less or equal than $n$. So, $p_{2}=2$, $p_{3}=3$, $p_{4}=3$, $p_{5}=5$, $p_{6}=5$ and so on.

Then, there exists the limit $\lim_{n}\frac{p_{n}}{n}$?

Many thanks in advances for your comments.

Answer 1

This question is closely related to the prime gap problem. The notation is slightly different: $p_n$ on Wikipedia is the $n^{th}$ prime while $g_n$ is the gap after that prime. I will stick to this notation, and use $a_n$ for the terms in your sequence to avoid confusion.

Wikipedia mentions that the quotient between the prime gap and the prime preceding the gap goes to 0: $$\lim_{n\to\infty}\frac{g_n}{p_n}=0.$$ Later I will use the prime following each gap. It follows trivially that also $\lim_{n\to\infty} g_{n-1} / p_{n}=0.$

Back to your problem. If the limit $\lim_{n \to \infty} a_n$ exists, it has to be 1 (since the value 1 is attained when $n$ is prime). The largest deviation between $a_n$ and 1 occurs when $n+1$ is prime. Consider the subsequence of points where the largest gaps occur, i.e., the subsequence $(a_{p_n-1})_{n=1}^\infty$. We have $a_{p_n-1} = (p_n-g_{n-1}) / (p_n-1)$, so $\lim_{n\to\infty} a_{p_n-1} = 1$. Therefore also $\lim_{n\to\infty} a_n = 1$

Answer 2

I will denote the number of primes less than or equal to $n$ a $\pi(n)$, the $n$th prime as $p_n$, and the largest prime less than or equal to $n$ as $q_n=p_{\pi(n)}$.

Clearly $1 \geq \frac{q_n}{n} \geq \frac{p_{\pi(n)}}{p_{\pi(n)+1}}$. Since $\lim_{n\rightarrow\infty}\frac{p_n}{p_{n+1}}=1$, we have that $\lim_{n\rightarrow\infty}\frac{q_n}{n}=1$ by the squeeze theorem.