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This is a question out of curiosity, while looking at the Firoozbakht's conjecture. It might not be research related, but as usual, I am not really sure if a question ever is research related or not, so excuses in advance if it is not:

The Firoozbakht's conjecture is:

$$\forall n \ge 1: p_{n+1}^{\frac{1}{n+1}} < p_{n}^{\frac{1}{n}}$$

where $p_k$ denotes the $k$-th prime number.

Taking logarithms on both sides we get:

$$\forall n \ge 1: \log(p_n) < \log(p_{n+1}) < (1+\frac{1}{n}) \log(p_n)$$

Dividing by $\log(p_n)$ we arrive at:

$$\forall n \ge 1: 1 < \frac{\log(p_{n+1})}{\log(p_n)} < 1+\frac{1}{n} $$

Summing the last inequalities for $k=1,\cdots,n$ we get:

$$\forall n \ge 1: n = 1+\cdots+1 < \sum_{k=1}^n \frac{\log(p_{k+1})}{\log(p_k)} < n+H_n$$

where $H_n$ denotes the $n-th$ Harmonic number: $H_n = \sum_{k=1}^n \frac{1}{k}$.

Dividing by $n$ on all sides of the inequalities, and taking the limit $n \rightarrow \infty $ we find:

$$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{\log(p_{k+1})}{\log(p_k)}=1$$

Forgetting how we arrived at this conclusion, we might ask:

  1. Can this last limit be proven with the help of the prime number theorem?

  2. Is there any known result, such for example the prime number theorem or something else in analytic number theory, to which the last limit is equivalent to, provided it can be proven.

Thanks for your help.

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    $\begingroup$ For (1), yes, by the prime number theorem $\log p_k \sim \log k$, and the limit you ask about follows immediately. In fact you can get away with something weaker than PNT, even e.g. Chebyshev's estimates suffice. $\endgroup$ Commented Aug 21, 2024 at 17:21

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Notice that $$ 1\leq \frac{\log p_{n+1}}{\log p_n}=1+\frac{\log(p_{n+1}/p_n)}{\log p_n}\leq 1+\frac{p_{n+1}-p_n}{p_n\log p_n}. $$ The last inequality is due to $\log(1+x)\leq x$ for $x\geq 0$. Now, for any $m$ we have $$ m\leq \sum_{k=m}^{2m-1}\frac{\log p_{k+1}}{\log p_k}\leq m+\sum_{k=m}^{2m-1}\frac{p_{k+1}-p_k}{p_k\log p_k}\leq $$ $$\leq m+\frac{1}{p_m\log m}\sum_{k=m}^{2m-1}(p_{k+1}-p_k)=m+\frac{p_{2m}-p_m}{p_m\log m}. $$ If we use PNT, we see that $p_{2m}=O(p_m)$, hence the last term is $O(1/\log m)$. Hence, if $2^M\leq n<2^{M+1}$, we obtain $$ n\leq \sum_{k=1}^{n}\frac{\log p_{k+1}}{\log p_k}\leq n+\sum_{m=1}^{M}O\left(\frac{1}{\log 2^m}\right)=n+O(\log\log n), $$ which is much stronger than we get from Firoozbakht's conjecture alone.

Note now that we only used that $p_{2m}=O(p_m)$, which is much weaker than the PNT, so I suspect that if your formula is equivalent to some statement about primes, that statement should be very weak compared to known estimates.

EDIT: To illustrate my point further, suppose that we only know that $p_m\geq m$ and $p_m=O(c^m)$ for some constant $c>1$. Then for large $x$

$$ \sum_{k\leq x}\frac{\log(p_{k+1}/p_k)}{\log p_k}\leq \frac{1}{\log 2}\sum_{k\leq M}\log(p_{k+1}/p_k)+\frac{1}{\log M}\sum_{M<k\leq x}\log(p_{k+1}/p_k) $$ for any $M\geq 2$. The first sum evaluates to $\log(p_{M+1}/2)=O(M)$ and the second is $\log(p_{[x]+1}/p_{M+1})=O(x)$, so we get $O(M+x/\log M)$ and choosing $M=\sqrt{x}$ we get the error term $O(x/\log x)$, which is enough to get the asymptotic formula.

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  • $\begingroup$ This is because Firoozbakht's conjecture is about a kind of worst-case scenario: typically, $p_{n+1}-p_n\approx \log p_n$. $\endgroup$ Commented Aug 21, 2024 at 17:59
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    $\begingroup$ @LSpice, of course not, it was just a force of habit. Corrected, thanks! $\endgroup$ Commented Aug 22, 2024 at 1:58
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    $\begingroup$ @mathoverflowUser, your formula sums from $m$ to $2m+1$, the sum in consideration is from $m$ to $2m$. The inequality holds because we sum $v_ku_k$ with positive $v_k,u_k$ and estimate it by $\max u_k \cdot\sum v_k$, where $v_k=p_{k+1}-p_k$, $u_k=\frac{1}{p_k\log p_k}$. $\endgroup$ Commented Aug 22, 2024 at 2:01
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    $\begingroup$ @mathoverflowUser you are right, I now sum form $m$ to $2m-1$ and get $m$ summands instead of $m+1$. The $2^{M-1}$ typo is also corrected $\endgroup$ Commented Aug 22, 2024 at 2:59
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    $\begingroup$ Thanks for your answer. I corrected a small typo. $\endgroup$ Commented Aug 22, 2024 at 3:10

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