Is $\sum_{k=1}^{n-1} \frac{1}{k}\frac{1}{n-k} \sim \frac{\log{n}}{n}$? I asked a similar question some days ago, but in a more general form that perhaps turned it in a too uninteresting question, so I'm asking it again in a more friendly form.
It is true that
$$\sum_{k=1}^{n-1} \frac{1}{k}\frac{1}{n-k} \sim \frac{\log{n}}{n} ?$$
I kind explained my intuition in the similar question that I linked above. Perhaps the only thing I tried so far was summation by parts, but it seems ineffective. I'm looking for a complete answer, but any kind of tips will be fairly welcome.
 A: $$\sum_{k=1}^{n-1} \frac{1}{k}\frac{1}{n-k} = 
\frac{1}{n}\sum_{k=1}^{n-1} \left(\frac{1}{k}+\frac{1}{n-k}\right) = 2\frac{H_{n-1}}{n}$$
where, $\displaystyle H_{n} = \sum\limits_{k=1}^{n} \frac{1}{k}$ is the $n$-th Harmonic Number. 
It is known that $H_n = \log n + \gamma +O(1/n)$.
A: As an alternate answer (though less exact), we can use the integral test of comparison. In general, if $f(x)$ is positive and decreasing, then 
$$ \sum_{k = 1}^n f(k) \approx \int_1^n f(x) dx,$$
where we could be explicit about error bounds if we wanted to. Here, we are looking at
$$ f(x) = \frac{1}{x(n - x)},$$
which is symmetric about the halfway point $x = n/2$. So although it only decreases for $x \in [1, n/2)$, by symmetry we can approximate the sum over the first half and then multiply by $2$. In other words,
$$ \sum_{k = 1}^n \approx 2 \int_1^{n/2} \frac{1}{x(n - x)}dx = \frac{2}{n} \log n.$$
This is only the leading term of sciona's answer, but I'm always pleased when we can approach things using elementary tools I teach in my first or second semesters of calculus.
