# Can $\{n!\log n\}$ converge to $1$?

Since $H_n-\log n\to\gamma$ is it correct to deduce that, if $\gamma$ were a rational $a/b$, then $$\lim_{n\to \infty }\{n!H_n-n!\log n\}=\lim_{n\to \infty }\{-n!\log n\}=1-\lim_{n\to \infty }\{n!\log n\}=0$$? If the implication is correct, what is known about $\{n!\log n\}$?

• $n! \log n$ grows so fast that I don't think currently anything can be said about its fractional part... But nice idea! Fourier's proof that $e$ is irrational does something similar, after all. – punctured dusk Jul 1 '18 at 14:18
• Excellent experimentation! Unfotrunately, your implication isn't correct, though mostly for somewhat subtle convergence issues. The fact that $H_n - \log n \to \gamma$ means there is some rate of convergence (which may be quite slow). You are trying to understand something like the fractional part of this error term after it's multiplied by $n!$, and $n!$ is gigantic. Thus any approach of this sort should not succeed. – davidlowryduda Jul 1 '18 at 14:30
• @ barto, @davidlowryduda Thank you both! – Richard Jul 1 '18 at 14:45

There appear to be two serious errors of thought here.

The first is that $\{ \cdot \}$ is not a continuous function. So, the fact that $a_n \to L$ does not mean you can clude $\{ a_n \} \to \{ L \}$. This would be true if $L$ wasn't an integer, but you are specifically interested about the case where it is.

The second problem is that you make absolutely no attempt to control the error of the approximation. In asymptotic notation, that $H_n - \log n \to \gamma$ means

$$H_n - \log n = \gamma + o(1)$$

multiplying by $n$ gives

$$n! H_n - n! \log n = \gamma n! + o(n!)$$

The problem is the error term $o(n!)$ is much too large for you to say anything about the fractional part of the right hand side.

• You are totally right, thank you – Richard Jul 1 '18 at 14:46