A weaker version of the Andrica's conjecture

Question

Andrica's conjecture states that: For every pair of consecutive prime numbers $p_{k}$ and $p_{k+1}$, we have :

$$\sqrt{p_{k+1}}-\sqrt{p_{k}}<1\quad\quad \color{#2d0}{\text{(1.)}}$$

I know that is statement is not yet proved. But I am asking on a weaker version:

Show that inequality $\color{#2d0}{\text{(1.)}}$ holds true for infinitely many indices $k$.

This is possible for the Legendre conjecture. See: Show that there exist infinitely many indices $k$ such that the intervals $[k^2,(k+1)^2]$ contain at least one prime number

Answer 1

Assuming the contrary it follows (induction) that there exists $A$ such that $\sqrt{p_k}> k-A$ for all $k$ and this would imply $\sum \frac{1}{p_k} < \infty$, contradiction.

Answer 2

It's easy to show that (given $p_n \sim n\ln{n}$) $$\color{red}{\frac{\sum\limits_{k=2}^n\left(\sqrt{p_k}-\sqrt{p_{k-1}}\right)}{n-1}}= \frac{\sqrt{p_n}-\sqrt{p_{1}}}{n-1}=\\ \frac{\sqrt{p_n}}{n-1}-\frac{\sqrt{p_{1}}}{n-1}=\\ \sqrt{\frac{p_n}{n\ln{n}}\cdot\frac{n\ln{n}}{({n-1)^2}}}-\frac{\sqrt{p_{1}}}{n-1}\sim \\ \sqrt{\frac{n\ln{n}}{({n-1)^2}}}\color{red}{\to 0}, n\to\infty$$ And now, if we assume that $\sqrt{p_k}-\sqrt{p_{k-1}}<1$ for a finite number of $k$'s, say $K$, then from some $n$ onwards
$$\color{blue}{\frac{\sum\limits_{k=2}^n\left(\sqrt{p_k}-\sqrt{p_{k-1}}\right)}{n-1}}=\\ \frac{\sum\limits_{\sqrt{p_k}-\sqrt{p_{k-1}}\geq1}\left(\sqrt{p_k}-\sqrt{p_{k-1}}\right)}{n-1}+ \frac{\sum\limits_{\sqrt{p_k}-\sqrt{p_{k-1}}<1}\left(\sqrt{p_k}-\sqrt{p_{k-1}}\right)}{n-1}\geq\\ \frac{n-K}{n-1}+ \frac{\sum\limits_{\sqrt{p_k}-\sqrt{p_{k-1}}<1}\left(\sqrt{p_k}-\sqrt{p_{k-1}}\right)}{n-1}\sim \frac{n-K}{n-1}\color{blue}{\to1}, n\to\infty$$ which contradicts the previous limit.

Answer 3

Assume that

$$\sqrt{p_{k+1}} - \sqrt{p_{k}} \lt 1 \tag{1}\label{eq1A}$$

only holds for a finite number of indices $k$. Thus, there's some integer $n_0$ where for all $k \ge n_0$ you have

$$\sqrt{p_{k+1}} - \sqrt{p_{k}} \ge 1 \tag{2}\label{eq2A}$$

Move $\sqrt{p_{k}}$ to the right side and, since they're both positive quantities, square both sides, and make some other adjustments to get

$$\begin{equation}\begin{aligned} \sqrt{p_{k+1}} & \ge \sqrt{p_{k}} + 1 \\ p_{k+1} & \ge p_{k} + 2\sqrt{p_{k}} + 1 \\ p_{k+1} - p_{k} & \ge 2\sqrt{p_{k}} + 1 \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

However, the Prime Number Theorem states the difference between consecutive primes is, asymptotically on average, the natural logarithm of the number (e.g., Wikipedia's "Prime Gap" page's Upper bounds section's second paragraph explicitly says "The prime number theorem, proven in 1896, says that the average length of the gap between a prime $p$ and the next prime will asymptotically approach $\ln(p)$ for sufficiently large primes"). Thus, you get

$$p_{k+1} - p_{k} \sim \ln(p_{k}) \tag{4}\label{eq4A}$$

However, for even fairly small primes, you have $2\sqrt{p_{k}} + 1 \gg \ln(p_{k})$. You can't have \eqref{eq3A} holding for all $k \ge n_0$ when the average value is so much less, so the original assumption of finiteness must be incorrect.

Another way to see this is sum the inequalities in \eqref{eq3A} to some $N - 1 \gg n_0$ to get

$$\begin{equation}\begin{aligned} \sum_{k=n_0}^{N-1}(p_{k+1} - p_{k}) & \ge \sum_{k=n_0}^{N-1}(2\sqrt{p_{k}} + 1) \\ p_{N} - p_{n_0} & \ge \sum_{k=n_0}^{N-1}(2\sqrt{p_{k}} + 1) \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

However, the PNT says $p_{N} \sim N\ln(N)$, which is much less than the right hand side of \eqref{eq5A}. As such, the original assumption must be wrong and, thus, there are an infinite number of indices $k$ for which \eqref{eq1A} holds.