Conditional Bounds for Prime Gaps with Applications

We posit that $d_n^2<2p_{n+1}$ holds for all $n\ge 1$, where $p_n$ represents the $n$th prime and $d_n$ stands for the $n$th prime gap i.e. $d_n:=p_{n+1}-p_n$. Then, the presence of a prime between successive perfect squares, as well as the validity of ${\mathrm{\Delta }}_n:=\sqrt{p_{n+1}}-\sqrt{p_n}<1$ are derived. Next, $\pi (x)$ being the number of primes $p$ up to $x$, we deduce $\pi (n^2-n)<\pi (n^2)<\pi (n^2+n)$ $(n\ge 2)$. In addition, a proof of $\pi ((n+1{)}^k)-\pi (n^k)\ge \pi (2^k)$ \ $(k\ge 2,n\ge 1)$ is worked out. The vanishing nature of ${\mathrm{\Delta }}_n$ as $n$ goes to infinity is set, and used afterwards to achieve both ${\displaystyle \underset{n\to \mathrm{\infty }}{\lim }{d}_n\mathrm{/}\sqrt{p_n}=0}$ and the twin prime conjecture. Also, question about the estimate $p_n<2j_n^2(n\ge 6)$, where $j_n$ counts the twin prime pairs up to $p_n$, is raised. Finally, we put forward the conjecture that any rational number $r$ $(0\le r\le 1)$ represents an accumulation point of the sequence ${(\{\sqrt{p_n}\})}_{n\ge 1}$, where $\{x\}$ acts for the fractional part of $x$.