TITLE: A weaker version of the Andrica's conjecture
QUESTION [9 upvotes]: 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

REPLY [6 votes]: 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.