Prove that $\lim_{n\rightarrow \infty} \frac{\log_{10}\lfloor\text{Denominator of } H_{10^n}\rfloor+1 }{10^n}=\log_{10} e$

Question

In short, my question is asking to prove that the $$\lim_{n\to\infty}\frac{\text{number of digits in the denominator of} \sum_{k=1}^{10^n} \frac 1k}{10^n}=\log_{10} e$$ I know that the number of digits in a number is $\lfloor \log_{10} n\rfloor +1$ and that the Harmonic numbers are given by $\gamma+\psi_0 (n+1)$, where $\psi_0 (x)=\frac{\Gamma'(x)}{\Gamma(x)}$ but I don't see how to find the number of digits in the denominator of $\psi_0(n+1)+\gamma$.

Context:From Wolfram MathWorld,

"The numbers of digits in the denominator of $H_{10^n}$ for $n=0, 1, \ldots$ is given by $1, 4, 40, 433, 4345, 43450, 434110, 4342302, 43428678, \ldots$ (OEIS A114468). These digits converge to what appears to be the decimal digits of $\log_{10}e=0.43429448\ldots$ (OEIS A002285)."

Answer 1

The denominator of the $M$th harmonic number, before cancelling out any common factors with the numerator, is $\mathop{\rm lcm}[1,\dots,M]$; by the prime number theorem, we know that $\ln \mathop{\rm lcm}[1,\dots,M] \sim M$, and hence the number of digits of $\mathop{\rm lcm}[1,\dots,M]$ is approximately $M/\ln10 = M\log_{10}e$.

The remaining step would be to show that there isn't too much cancellation between the numerator and denominator of $M$. I'm willing to believe that that is an open problem, though I don't know for sure.

Answer 2

Let $D(n)$ be the denominator of $H(n)$ and let $L(n) = LCM(1,2,\ldots,n)$. It is easy enough to give bounds good enough to show that $$\lim_{n \to \infty} \frac{\log L(n)}{n} = \lim_{n \to \infty} \frac{\log D(n)}{n} = 1.$$ (All logs are base $e$.)

It is well known that $\lim_{n \to \infty} \frac{\log L(n)}{n}=1$, and clearly $D(n) \leq L(n)$, so the point is to provide lower bounds. Fix a large positive integer $k$. Choose $N$ large enough that $N>k^2$ and $N/k$ is large than any prime dividing the numerator of one of $H_1$, $H_2$, ..., $H_k$. We will bound $H_n$ for $n>N$.

Let $p$ be a prime between $n/k$ and $n$. Then the summands in $H_n$ whose denominator is divisble by $p$ are $1/p$, $1/(2p)$, $1/(3p)$, ..., $1/(\ell p)$ for some $\ell \leq k$. They add up to $H_{\ell}/p$. Since $p>n/k$ and $\ell \leq k$, we know that $p$ does not divide the numerator of $H_{\ell}$. Set $H_{\ell} = a/b$. Then $H_n$ is of the form $(a/b)/p + c/d=(ad+pcb)/(pbd)$ where $p$ does not divide $a$ or $d$. So $p$ divides the denominator of $H_n$.

We see that $$\log D(n) \geq \sum_{n/k<p<n} \log p = \vartheta(n) - \vartheta(n/k) \sim n\cdot (1-1/k).$$ Here $\vartheta(x)$ is Chebyshev's notation for $\sum_{p < x} \log p$ and the asymptotic relation is from the prime number theorem. So $\lim \inf_{n \to \infty} \frac{\log D(n)}{n} \geq 1-1/k$. Since $k$ was arbitrary, $\lim \inf_{n \to \infty} \frac{\log D(n)}{n} \geq 1$.

Examples suggest that $L(n)/D(n)$ is very small compared to $L(n)$: You can see some values here.