I checked some relations between primes, here $$1 and $$p_n$$ is the $$n$$th prime.

$$a) p_n^{1/3} - p_{n-1}^{1/3}<1/2$$

$$b) p_n^{1/n} - p_{n-1}^{1/n}<1/n$$

$$c) (\log p_n)^{1/2} - (\log p_{n-1})^{1/2} < 1/4$$

$$d) (\log p_n)^{1/n} - (\log p_{n-1})^{1/n} < 1/n^X, n\geq7,X=2$$

In $$d)$$ I tried to find a larger $$X$$, but I failed.

Maybe some will fail for a larger $$n$$. Are any of these know to be true? Also, what can be deduced from they?

• a) is true for sufficiently large $n$, by sufficient strengthenings of Bertrand's postulate. E.g., it is known that $[x, x+x^{0.525}]$ contains a prime for all sufficiently large $x$. Evaluating at $x=p_{n-1}$ gives $p_n^{1/3}-p_{n-1}^{1/3}\to 0$. Feb 6 at 3:54

The first three statements are true for $$n$$ sufficiently large. The fourth statement is equivalent to a very strong bound on prime gaps (which has a chance to hold but is right on the edge and hopeless to prove or disprove at present), while any $$X>2$$ produces a false bound for all $$n>n_0(X)$$.
Regarding a), a classical result of Ingham (1937) shows that $$p_n^{1/3} - p_{n-1}^{1/3}<\frac{p_n-p_{n-1}}{3p_{n-1}^{2/3}}=o(1).$$
Regarding b), the Prime Number Theorem gives that $$p_n^{1/n} - p_{n-1}^{1/n}\sim\frac{p_n-p_{n-1}}{n p_{n-1}}=o\left(\frac{1}{n}\right).$$
Regarding c), Bertrand's postulate shows for $$p_n>7$$ that $$(\log p_n)^{1/2}-(\log p_{n-1})^{1/2}<\frac{\log p_n-\log p_{n-1}}{2(\log p_{n-1})^{1/2}}<\frac{\log 2}{2(\log p_{n-1})^{1/2}}<\frac{1}{4}.$$ In fact the left-hand side tends to zero (as $$n\to\infty$$).
Regarding d), the Prime Number Theorem gives that $$(\log p_n)^{1/n}-(\log p_{n-1})^{1/n}\sim\frac{p_n-p_{n-1}}{n p_n\log p_n}\sim\frac{p_n-p_{n-1}}{n^2\log^2 n}.$$ Under some standard (but bold) conjectures the right-hand side stays below $$2/n^2$$, and perhaps it is even $$o(1/n^2)$$. So statement d) restricted to large $$n$$ is equivalent to some expected but hopelessly difficult upper bound on prime gaps. On the other hand, the above calculation shows that d) fails for any $$X>2$$ and $$n>n_0(X)$$.