This article on the harmonic series says that
$$\sum_{n=1}^k\,\frac{1}{n} \;=\; \ln k + \gamma + \varepsilon_k$$
where $$\varepsilon_k\sim\frac{1}{2k}$$
and this seems to generalise to $$\sum_{n=1}^k\,\frac{1}{\ln(x)n} \;=\; \log_x k + \frac{\gamma}{\ln(x)} + \varepsilon_k $$
Why is this the case?
Consider the area below $1/x$ between 1 and $n$, and compare the curve to the "staircase" $1/k$. The area under the staircase is your sum, the area under the curve is just $\ln n$. If you look at the curve $1/(x - 1)$, it is above the staircase, an approximation from above to the staircase area is $1+\int_2^n \frac{d x}{1-x} = 1 + \ln (n - 1)$. Averaging both areas gives a rough estimate of $\ln n + 1/2 + \ln \frac{n - 1}{n} \approx \ln n + 1/2 - 1/n$ for the sum.
The image shows the relationship between a continuous function (green) and the staircase (red) described.