Comparing the harmonic sum with the logarithm

Two comparisons. For $k\ge1$ and $x\in[k,k+1]$, $\frac1{k+1}\le\frac1x\le\frac1k$. Integrating over $[k,k+1]$ (length $1$):
$$\frac1{k+1}\le\int_k^{k+1}\frac{dx}x\le\frac1k.$$

Lower bound. Using the right inequality and summing over $k=1,\dots,n$:
$$H_n\ge\int_1^{n+1}\frac{dx}x=\ln(n+1).$$

Upper bound. Using the left inequality, $\frac1k\le\int_{k-1}^k\frac{dx}x$ for $k\ge2$. Summing over $k=2,\dots,n$:
$$H_n-1\le\int_1^n\frac{dx}x=\ln n.$$

Consequences. Since $\ln(n+1)\to+\infty$, $H_n\to+\infty$. Dividing by $\ln n$,
$$\frac{\ln(n+1)}{\ln n}\le\frac{H_n}{\ln n}\le\frac1{\ln n}+1,$$
and both bounds tend to $1$.

Remark. The difference $H_n-\ln n$ is decreasing and bounded below, so it converges to a constant, the Euler–Mascheroni constant $\gamma\approx0.5772$.