Let $a,k,N$ be positive integers with $k\leq n-a$, and $0< a<n$. Let $$\theta(k) = \prod_{j=0}^{k-1}\left(1-\frac{a}{n-j}\right)$$ It is clear that $\theta(k) \leq \prod_{j=0}^{k-1} (1-\frac{a}{n})= (1-\frac{a}{n})^k$.
Then let's consider the sum $$\begin{align} S = \sum_{k=0}^{n-a} \frac{\theta(k)}{k} &< \sum_{k=0}^{n-a} \frac{(1-\frac{a}{n})^k}{k} \\&= \sum_{k=0}^{\infty} \frac{(1-\frac{a}{n})^k}{k} - \sum_{k=n-a+1}^{\infty} \frac{(1-\frac{a}{n})^k}{k} \\&< \log\left(\frac{n}{a}\right) \qquad \text{using Taylor series} \end{align} $$
I'm interested in the order of the error term in approximating $S$ by $\log(n/a)$. I'm at a loss on how to handle the first approximation (which I used to simplify the product). The second approximation I can handle using the error term of Taylor series.
$S = \sum \limits_{k=1}^{n-a} \frac{\theta(k)}{k} = \sum \limits_{k=1}^{n-a} \frac{\prod \limits_{j=0}^{k-1} (1-\frac{a}{n-j})}{k} = \sum \limits_{k=1}^{n-a} \frac{\prod \limits_{j=0}^{k-1} \frac{n-a-j}{n-j}}{k} = \sum \limits_{k=1}^{n-a} \frac{\frac{(n-a)(n-a-1) \cdots (n-a-k+1)}{(n)(n-1)(n-2)\cdots (n-k+1)}}{k} = \sum \limits_{k=1}^{n-a} \frac{\frac{(n-k)! (n-a)! }{n! (n-a-k)!}}{k} $
For $a=0$ => $\sum \limits_{k=1}^{n} \frac{1}{k} = H_n$
For $a=1$ => $\sum \limits_{k=1}^{n-1} \frac{1}{k}-\frac{1}{n} = H_{n-1}+\frac{1}{n}-1$
For $a=2$ => $H_{n-2}+\frac{1}{n-1}+\frac{1}{n}-\frac{3}{2}$
You see a pattern (this pattern can be proved by induction with the above as base cases).
The pattern is $H_{n-a} - H_a +\sum \limits_{j=0}^{a-1} \frac{1}{n-j}$
Now what is $\sum \limits_{j=0}^{a-1} \frac{1}{n-j} = \frac{1}{n}+\frac{1}{n-1}+\frac{1}{n-2}+\cdots +\frac{1}{n-a} = \frac{1}{n}+\frac{1}{n-1}+\frac{1}{n-2}+\cdots +\frac{1}{n-a}-\frac{1}{n-a}+ \frac{1}{n-a-1} - \frac{1}{n-a-1} +\cdots +\frac{1}{1}-\frac{1}{1} = H_n-H_{n-a}$.
So the final answer is $H_{n-a}-H_a +H_n-H_{n-a} = H_n-H_a $
$\ln(\frac{n}{a})+\frac{1}{2n}-\frac{1}{2a} -\frac{1}{12 n^2} < H_n-H_a < \ln(\frac{n}{a})+\frac{1}{2n}-\frac{1}{2a} + \frac{1}{12a^2}$.
So in the end your approximation was very very close to the exact answer.