TITLE: Closed form for the partial sum $\sum\limits_{k = 1}^n \frac{\ln k}k$
QUESTION [6 upvotes]: I'd like to find a closed form for this partial sum:
$$\sum\limits_{k = 1}^n \frac{\ln k}k$$
Using the properties of the logarithms, I converted the above into
$$\ln\left(\prod_{k = 1}^n k^{1/k}\right),$$
but I don't know how to proceed.
Is this the right direction? If not, what can I do?

REPLY [4 votes]: We have, by partial summation:
$$ \sum_{k=1}^{n}\frac{H_k}{k} = H_n^2 - \sum_{k=1}^{n-1}\frac{H_k}{k+1} $$
hence it follows that:
$$ \sum_{k=1}^{n}\frac{H_k}{k} = \frac{H_k^2+H_k^{(2)}}{2} \tag{1}$$
and since:
$$ H_n = \log n +\gamma +\frac{1}{2n}+O\left(\frac{1}{n^2}\right)\tag{2} $$
it follows that:
$$ \sum_{k=1}^{n}\frac{\log k}{k} = O(1)+ \sum_{k=1}^{n}\frac{H_k-\gamma}{k}=O(1)+\frac{1}{2}H_k^2-\gamma H_k=\frac{1}{2}\log^2 n+O(1).\tag{3}$$