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)."