# Do the ratios of successive primes converge to a value less than 1?

I think it's a pretty straightforward question.

Does $\lim_{n \to \infty}{\frac{p_n}{p_{n+1}}} < 1?$

***$p_n$ denotes the nth prime.

Since the average gap increases between successive primes by the prime number theorem, wouldn't this be the case?

• The limit exists and is equal to $1$. – André Nicolas Aug 16 '14 at 19:05
• @JonasMeyer, sorry. I was counting something else. You are correct. I removed my previous comment. – ReverseFlow Aug 16 '14 at 19:30

By the prime number theorem,

$$\lim_{n\to\infty} \frac {p_n} {n \log n} = 1$$

Replacing $n$ by $n+1$,

$$\lim_{n\to\infty} \frac {p_{n+1}} {(n+1) \log (n+1)} = 1$$

Using limit arithmetic,

$$\lim_{n\to\infty} \frac {p_n} {p_{n+1}} \frac {(n+1)\log(n+1)}{n \log n} = 1$$

However, it is elementary that

$$\lim_{n\to\infty} \frac {(n+1)\log(n+1)}{n \log n} = 1$$

and we conclude (again using limit arithmetic - dividing the last two results):

$$\lim_{n\to\infty} \frac {p_n}{p_{n+1}} = 1$$

• Could you please explain the reason for the downvote? – Yoni Rozenshein Aug 16 '14 at 19:27

Edit: As pointed out in a comment, the previous version of this answer fell a little short of what it could have said.

As noted in The prime number theorem and the nth prime, if we write $p_n$ for the $n$th prime number, then $p_n \sim n \ln n.$ That is, $\lim_{n \rightarrow \infty}\frac{p_n}{n \ln n} = 1.$ But

$$\lim_{n \rightarrow \infty}\frac {n \ln n}{(n + 1) \ln (n + 1)} = 1.$$

So we have

$$\lim_{n \rightarrow \infty}\frac {p_n}{p_{n+1}} = \lim_{n \rightarrow \infty} \left( \frac {p_n}{n \ln n} \cdot \frac{n \ln n}{(n + 1) \ln (n + 1)} \cdot \frac{(n + 1) \ln (n + 1)}{p_{n+1}} \right) = 1.$$

At http://mathworld.wolfram.com/PrimeGaps.html you can find more about what is known about the differences between successive primes. For example, if $g(N)$ is defined as the least upper bound of the difference between any two consecutive primes $p_n$ and $p_{n+1}$ where $p_{n+1} < N,$ then it is conjectured that $g(N) \sim (\ln N)^2.$

• You're right, I overlooked the obvious stronger conclusion. I suppose I got distracted by the conjectured $(\ln N)^2$ asymptotic upper bound. I have edited the answer since its previous version was somewhat misleading, although the correct answer that was posted in the meantime takes precedence. – David K Aug 17 '14 at 1:21