The Loga-rhythm of Harmony

See the comments here .

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Notice that $\frac{1}{f(H_{k})}$ is the denominator of $H_{k}$ , since $H_{k}$ is always rational. To make things convenient, let's call $1/f(x) = d(x)$ for denominator.

Now, what do we usually do with weird-looking limits? Take their logarithm, of course!

$\log_{10} \lim_{n \rightarrow \infty} f(H_{10^{n}})^{-1/10^{n}} = \lim_{n \rightarrow \infty} \frac{\log_{10} d(H_{10^{n}})}{10^{n}}$

Now, we know that $\lceil \log_{10} N \rceil$ is the number of decimal digits of $N \neq 10^{n}$ for some positive integral $n$ . Since $d(H_{10^{n}}) \rightarrow \infty$ as $n \rightarrow \infty$ , we can just ignore the ceiling function. If $d_{\#}(N)$ is the number of decimal digits of $N$ , then

$\lim_{n \rightarrow \infty} \frac{\log_{10} d(H_{10^{n}})}{10^{n}} = \lim_{n \rightarrow \infty} \frac{d_{\#}(H_{10^{n}})}{10^{n}} = \log_{10} e$

which is a consequence of this sequence 's properties.

So, we can finally conclude

$\lim_{n \rightarrow \infty} f(H_{10^{n}})^{-1/10^{n}} = 10^{\log_{10} e} = \boxed{e = 2.718\ldots}$