$n>1$, $p_n$ is the prime number nth then $p_{n+1}^{n^2} > p_n^{n^2+1}$ (conjecture)

Question

Let $p_n$ is prime number nth:

If n>1 then: $$p_{n+1}^{n^2} > p_n^{n^2+1}$$

I checked the conjecture above true for first fifty million primes.

Could You give your remark, reference, or your proof of conjecture above?

Answer 1

$p_{n+1}^{n^2} > p_n^{n^2+1}$ is the same as $p_{n+1} \gt p_n^{1+1/n^2} =p_n\cdot p_n^{1/n^2} $.

Since $p_n \approx n \ln n$,

$\begin{array}\\ p_n^{1/n^2} &\approx (n \ln n)^{1/n^2}\\ &=e^ {\ln(n \ln n)/n^2}\\ &=e^ {(\ln n+\ln \ln n)/n^2}\\ &<e^ {2\ln n/n^2}\\ &\approx 1+2\ln n/n^2\\ \text{so}\\ p_n\cdot p_n^{1/n^2} &<p_n\cdot (1+2\ln n/n^2)\\ &=p_n+2p_n\ln n/n^2\\ &\approx p_n+2(n \ln n)\ln n/n^2\\ &= p_n+2\ln^2 n/n\\ &< p_n+1 \qquad\text{for large } n\\ \end{array} $

since $\ln^2 n/n \to 0$.

If your original inequality is $p_{n+1}^{n} > p_n^{n+1} $, this becomes $p_{n+1}^{1/(n+1)} > p_n^{1/n} $ which contradicts Firoozbakht's conjecture (see https://en.wikipedia.org/wiki/Prime_gap#Conjectures_about_gaps_between_primes for this and other conjectures).