Apparent link between the zeta function and $\ln(10)$

Question

Why does $$\sum_{i=10^x}^{10^{(x+1)}-1} \frac{1}{i}$$ seem to tend towards $\ln(10)$ as $x $ increases?

Answer 1

$$\sum_{i=10^x}^{10^{(x+1)}-1} \frac{1}{i}=H_{10^{x+1}-1}-H_{10^x-1}=\log10^{x+1}-\log10^x+o(1)=\log10+o(1)$$ where $H_n$ denotes the n-th harmonic number