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This question is from Nagura's 1952 paper on primes in short intervals. He uses $\psi(x) = \sum_{m=1}^{\infty}\phi(\sqrt[m]{x})$, in which $\phi(x) = \sum_{ p \leq x }\ln p$. Using close arguments about size he obtains an upper bound for $\psi(x)$ as $\psi(x)< 1.086x$ for $x>0$. One thing he does not do to improve the bound is the following:

1) If $\psi(x) < 1.086x$ then $0.927 \psi(x) = \psi(x) (1 - 0.073) < 1.0053x$, and

$\psi(x) - \psi(x^{(0.1)}< \psi(x) - 0.073 \ \psi(x) = \psi(x) - \psi(x^{.073})< 1.0053x $

Then find an m such that $\psi(\frac{x}{m}) > \psi(x^{\frac{1}{10}}) $ for x sufficiently large, because then we would have $\psi(x) - \psi(\frac{x}{m})< \psi(x) - \psi(x^{\frac{1}{10}})< 1.0053x$ and could use a convenient trick to get rid of the negative term in the inequality.

2) Let x = $k^{10} $ so $\psi(x^{\frac{1}{10}}) = \psi(k). $ Now let $m = k^5 $ so that $\psi(\frac{x}{m}) = \psi(k^2) > \psi(k)$. Now double x so x =$ 2k^{10}$ and $\psi(x^{\frac{1}{10}}) = \psi(2^{\frac{1}{10}}k)$ and so $\psi(\frac{x}{m}) > \psi(x^{\frac{1}{10}})$ for $ k^{10}, 2k^{10}$, and so on.

3) Choose k to be 3 so that we get the benefit of the trick:

$\psi(x) - \psi(\frac{x}{243}) < \psi(x) - \psi(x^{\frac{1}{10}}) < 1.0053x$. Since the inequality is true for $0 < x < 243$ (because it it true for $\psi(x) < 1.086x$) , we can substitute $\frac{x}{243}$ for x, then $\frac{x}{243^2}$, and so on, obtaining:

$\psi(x) - \psi(\frac{x}{243}) < 1.0053x$

$\psi(\frac{x}{243}) - \psi(\frac{x}{243^2}) < 1.0053 \frac{x}{243}$

$\psi(\frac{x}{243^2}) - \psi(\frac{x}{243^3}) < 1.0053 \frac{x}{243^2}$ and so on.

We add these, getting :

$\psi(x) < 1.0053 (x + \frac{x}{243} + \frac{x}{243^2}+ ...) < 1.0095 x$

Substituting the new upper bound into Nagura's final equation, we have:

$\phi(\frac{(n+1)x}{n}) - \phi(x) >$

$ 0.916 ( \frac{(n+1)x}{n}+\sqrt{x}+\sqrt[3]{x})- 6.954 - 1.0095(x + \sqrt{\frac{n+1}{n}x }+ \sqrt[3]{\frac{n+1}{n}x} + \sqrt[5]{ \frac{n+1}{n}x}) $

This expression becomes positive for up to n = 9 (at about x = 2557) and after n = 9 it fails to become positive. If this approach were correct we would say there is a prime between $\frac{10}{9}x $ and x It can't be correct because 1.0053 was somewhat arbitrary. I think maybe there is an error in (2)? Thanks.

Edit: 2) Let x = $k^{10} $ so $\psi(x^{\frac{1}{10}}) = \psi(k). $ Now let $m = k^5 $ so that $\psi(\frac{x}{m}) = \psi(k^5) > \psi(k)$. Now double x so x =$ 2k^{10}$ and $\psi(\frac{x}{m}) = \psi(2k^5)$, $\psi(x^{\frac{1}{10}}) = \psi(2^{\frac{1}{10}}k)$ and so $\psi(\frac{x}{m}) > \psi(x^{\frac{1}{10}})$ for $ k^{10}, 2k^{10}$, and so on...

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up vote 2 down vote accepted

Yes, it does seem that there is an error in (2); if $x=k^{10}$, and $m=k^5$, then $x/m=k^5$, not $x/m=k^2$.

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I appreciate that you found this error but it only seems to strengthen the inequality (?)(see edit). If so there must be a more serious flaw in the argument. – daniel Jan 3 '12 at 10:54
True, but when I see something like $k^{10}/k^5=k^2$, I reckon it's OP's job to have a more careful look at the write-up and find the errors himself. – Gerry Myerson Jan 3 '12 at 14:48
Fair enough. (Have to add more characters...) – daniel Jan 3 '12 at 21:48

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