Prove that $\pi(n^2)-\pi(\frac{n^2+2n}{2})>0$

Question

I'd like to know if there's a better way to prove that:

$$\pi\left(n^2\right)-\pi\left(\frac{n^2+2n}{2}\right)>0$$

than using "There's always a prime in $(m-m^{23/42},m)$" by Iwaniec-Pintz:

(I put $m=n^2-1$, so $n^2-1-(n^2-1)^{23/42}$ must be $>\frac{n^2+2n}{2}$, true for $n>4$);

Thanks!

Answer 1

Using$$\frac{x}{\log\left(x\right)}<\pi\left(x\right)<1.25506\,\frac{x}{\log\left(x\right)}$$ if $x$ is sufficiently large, we have$$\pi\left(n^{2}\right)-\pi\left(\frac{n^{2}+2n}{2}\right)>\frac{n^{2}}{\log\left(n^{2}\right)}-1.25506\frac{n^{2}+2n}{2\log\left(\frac{n^{2}}{2}+n\right)}=n^{2}\left(\frac{1}{\log\left(n^{2}\right)}-\frac{1.25506}{\log\left(\left(\frac{n^{2}}{2}+n\right)^{2}\right)}-\frac{1.25506}{n\log\left(\frac{n^{2}}{2}+n\right)}\right)>0$$ if $n$ is sufficiently large because $\log\left(\left(\frac{n^{2}}{2}+n\right)^{2}\right)\sim\log\left(n^{4}\right)$ is bigger than $\log\left(n^{2}\right)$ as $n\rightarrow\infty$ Addendum: Note that is fundamental for this proof that $1.25506<2$. In fact the "complete" exponent of the second term is $$\frac{1}{\log\left(\left(\frac{n^{2}}{2}+n\right)^{2/1.25506}\right)}$$ but the reasoning works because is bigger than $1$.