Closed form for the partial sum $\sum\limits_{k = 1}^n \frac{\ln k}k$

Question

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?

Answer 1

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:

$$ \sum_{k=1}^{n}\frac{\log k}{k} = O(1)+ \sum_{k=1}^{n}\frac{H_k-\gamma}{k}=O(1)+\frac{1}{2}H_k^2-\gamma H_k=\frac{1}{2}\log^2 n+O(1).\tag{3}$$

Answer 2

It's asymptotically $O(\log^2 n)$, you can get it using Euler-Maclaurin formula. I don't think closed-form expression exists.

Answer 3

I am not sure that a closed form exists on the basis of elementary functions.

There is a closed form which involves the gives the Stieltjes constant as well as the generalized Stieltjes constant $$\sum\limits_{k = 1}^n \frac{\ln k}k=\gamma _1-\gamma _1(n+1)$$ what I am not sure you will enjoy.