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I am going through the proof by Jitsuro Nagura which shows that there is always a prime between $x$ and $\frac{6x}{5}$ where $x \ge 25$.

Nagura uses the following definitions:

$$\vartheta(x) = \sum_{p \le x} \log p$$

$$\psi(x) = \sum_{m=1}^{\infty}\vartheta(\sqrt[m]{x})$$

Then as part of the main theorem, he states:

$$\psi(x) - \psi(\sqrt{x}) - \psi(\sqrt[3]{x}) \ge \vartheta(x) \ge \psi(x) - \psi(\sqrt{x}) - \psi(\sqrt[3]{x}) - \psi(\sqrt[5]{x})$$

I've reviewed this inequality and am not clear on its justification. If someone could help explain how this inequality is proved, I would really appreciate

Thanks,

-Larry

Update: I think that I have figured out part of the inequality.

If I am doing my math right:

$$\psi(x) - \psi(\sqrt{x}) - \psi(\sqrt[3]{x}) - \psi(\sqrt[5]{x}) = $$

$$\vartheta(x) - \vartheta(\sqrt[6]{x}) + \vartheta(\sqrt[7]{x}) - \vartheta(\sqrt[10]{x}) + \vartheta(\sqrt[11]{x}) - \vartheta(\sqrt[12]{x}) + \vartheta(\sqrt[13]{x}) - \vartheta(\sqrt[15]{x}) + \vartheta(\sqrt[17]{x}) - \vartheta(\sqrt[18]{x}) + \vartheta(\sqrt[19]{x}) -\vartheta(\sqrt[20]{x}) +\vartheta(\sqrt[23]{x}) - \vartheta(\sqrt[24]{x}) + \vartheta(\sqrt[29]{x}) -2\vartheta(\sqrt[30]{x}) + \ldots \le $$

$$\vartheta(x) - \vartheta(\sqrt[6]{x}) + \ldots + \vartheta(\sqrt[30]{x}) + \ldots$$

By the properties of decreasing sequences of alternating signs:

$$\vartheta(x) - \vartheta(\sqrt[6]{x}) + \ldots + \vartheta(\sqrt[30]{x}) + \ldots \le \vartheta$$

Glorfindel
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Larry Freeman
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1 Answers1

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It's not too hard to see what's going on in these bounds. Fix an $x$ and let $a_k = \vartheta(\sqrt[k]{x})$. Then $\psi(x) = \sum_k a_k$ and more generally $\psi(\sqrt[n]x) = \sum_k a_{nk}$. Also as you note in your edit, the sequence $\{a_k\}$ is decreasing, albeit not strictly (in fact it is eventually zero).

So $\psi(x) - \psi(\sqrt[2]x) = \sum_{k~\text{odd}} a_k = a_1 + \sum_k a_{2k+1} \ge a_1 + \sum_k a_{3k} = \vartheta(x) + \psi(\sqrt[3]x).$

On the other hand, a careful calculation shows that $\sum_k a_{3k} + \sum_k a_{5k} \ge \sum_k a_{2k+1}$. I don't see a non-tedious way to prove this, but it is easy to tell that the first few terms on the left dominate those on the right, and since $\tfrac13 + \tfrac15 > \tfrac12$, the LHS has a higher density of terms than the RHS in the long run (the only question being how many initial terms it suffices to test).

An intuitive motivation for these inequalities comes from Möbius inversion, which gives the exact formula

$$\vartheta(x) = \sum_{k=1}^\infty \mu(n) \psi(\sqrt[n]x) = \psi(x) - \psi(\sqrt[2]x) -\psi(\sqrt[3]x) - \psi(\sqrt[5]x) + \psi(\sqrt[6]x) - \cdots$$

When $x$ is large enough, the size of any given term is greater than all of the terms to the right, so this behaves in some sense like a series approximation. If we truncate before the $-\psi(\sqrt[5]x)$ term we'd expect to get an overestimate (hence the first inequality), while if we truncate before the $+ \psi(\sqrt[6]x)$ term we'd expect to get an underestimate (hence the second inequality). What takes more care is establishing this for all $x$ large or small.

Erick Wong
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