Approximating $\frac{(n+1)H_n -n}{n^2}$ I have the following value
$$\frac{(n+1)H_n -n}{n^2}$$
But it is complicated to write, so I want to write a simple approximation to use it. I guess I can just write
$$\frac{(n+1)H_n -n}{n^2}\approx \frac{H_n-1}{n} \approx \frac{H_n}{n} \approx \frac{\ln n}{n}$$
Is this a good approximation, or is there a clearly better one? Thanks.

Very similar to @heropup answer and solution, we could use for
$$f(n) = \frac{(n+1)H_n - n}{n^2}$$
$$\frac{-1 + \gamma + \log n}{n} + \frac{1 + 2\gamma + 2 \log n}{2n^2} + \frac{5}{12n^3}\left(1-\frac{2310 n+559 } {50 \left(231 n^2+79 n+31\right) } \right) $$ which is equivalent to an $O\left(\frac{1}{n^8}\right)$ expansion.
For $n=10$, the relative error is $1.71\times 10^{-8}$%.
