I have been fooling around with logarithms and experimentally found a very strict inequality from above (which I haven't been able to improve further):
$$\ln n < H_{n-1}+\frac{1}{8(n-1)}+\frac{3}{8n}-\gamma \tag{1}$$
Here's the numerical check in Mathematica for $n=2 .. 2000$:
This is different from the usual asymptotic expansion of Harmonic numbers, and I'm not sure how to prove it. Any thoughts?
Maybe it's not true for larger $n$, so disproving it is also acceptable.
I would also like any sources for it, if it is already known (which I'm sure it is).
The usual asymptotics $$H_{n}\sim \ln {n}+\gamma +{\frac {1}{2n}}-\sum _{k=1}^{\infty }{\frac {B_{2k}}{2kn^{2k}}}=\ln {n}+\gamma +{\frac {1}{2n}}-{\frac {1}{12n^{2}}}+{\frac {1}{120n^{4}}}-\cdots$$
is of course better than (1), if we include the quadratic term, but worse if we only leave the linear term.
Mixing up those asymptotics up to the linear terms, I have found another inequality:
$$H_{n-1} < \frac{1}{2} ( \ln (n-1) + \ln n) +\frac{3}{16} \left( \frac{1}{n-1}-\frac{1}{n} \right) - \gamma \tag{2}$$
or:
$$H_{n-1} < \frac{1}{2} \ln n (n-1) +\frac{3}{16} \frac{1}{n(n-1)} - \gamma$$
Which is even more strict, and I bet difficult to prove.