Asymptotic of $\sum (H_n-H_{k-1})\ln(k)$ where $H_k$ are the harmonic numbers I want to find the asymptotic of $f(n)=$
\begin{equation}
\sum_{k=2}^n (H_n-H_{k-1})\ln(k)=\left(\frac{1}{2}+\cdots+\frac{1}{n}\right)\ln(2)+\left(\frac{1}{3}+\cdots+\frac{1}{n}\right)\ln(3)+\cdots+\frac{1}{n}\ln(n),
\end{equation}
up to the constant level. Using Abel's summation formula, let $a_k=H_n-H_{k-1}$, $A_k=\sum_{i=1}^k a_k=[u]\left(1+\frac{1}{[u]+1}+\cdots+\frac{1}{n}\right)$, and $\phi(x)=\ln(x)$. We have
\begin{align}
&\sum_{1<k\leq n}a_k\phi(k)=A(n)\phi(n)-A(1)\phi(1)-\int_1^n A(u)\phi'(u)du\\
=&n\ln n-\int_1^n \frac{[u]}{u}\left(1+\frac{1}{[u]+1}+\cdots+\frac{1}{n}\right)du.
\end{align}
How to proceed from here? It seems the integral part is not small.
 A: I think the best tool for these type of estimates (slow growing/monotone functions) is the quantitative integral test from Terry Tao's number theory notes: https://terrytao.wordpress.com/2014/11/23/254a-notes-1-elementary-multiplicative-number-theory/. See in particular Lemma 2, Exercise 7 and 10(a), and equations (8), (9), and the one above (14). In particular I will use: (all logarithms are natural)
$$H_n-H_{k-1}=\sum_{k\leq j\leq n} \frac 1j = \log n - \log k + O(\tfrac 1k),$$
$$\sum_{k=2}^n \log k = n \log n - n + O(\log(2+n)),$$
$$\begin{aligned}
\sum_{k=2}^n \log^2k &= \int_2^n \log^2 t dt + O(\log^2(2+n)) \\
&= (x\log^2x-2x \log x + 2x)\big|_1^n + O(\log^2(n+2))\\
&= n\log^2 n - 2n \log n + 2n + O(\log^2(n+2)).
\end{aligned}$$
and (Ex. 10(a)):
$$\sum_{k=2}^n \frac{\log k}k = \frac 12 \log^2n + O(1)$$
Thus,
$$\begin{aligned}
\sum_{k=2}^n (H_n-H_{k-1}) \log k &= \sum_{k=2}^n \big(\log n - \log k + O(\tfrac 1k)\big)\log k \\
&= \log n \sum_{k=2}^n \log k - \sum_{k=2}^n \log^2 k + \sum_{k=2}^n O(\tfrac 1k) \log k \\
&= \log n\big(n\log n - n + O(\log(2+n)\big) \\
&- \big(n\log^2n - 2n \log n + 2n + O(\log^2(n+2)\big)\\
&+O\Big(\sum_{k=2}^n \frac{\log k}k \Big)\\
&= n\log n - 2n + O(\log^2(2+n))
\end{aligned}$$
A: Using summation by parts:
Let $f_k = H_n- H_k$ and $G_k = \log k = g_{k+1} -g_k$,and $g_k = \sum_{j=2}^{k-1} \log j$
Then
$$ \sum_{k=2}^n f_k G_k = f_n g_{n+1} - f_2 g_2 - \sum_{k=3}^n g_k (f_k-f_{k-1}) = \sum_{k=3}^n \frac{g_k}{k} $$
because $f_n= g_2=0$
Now,
$$ g_{k+1} = k \log k - k +\frac12 \log k + \frac12 \log 2\pi + o(1)$$
$$ \frac{g_{k}}{k} = \log k  -1 + \frac32 \frac{\log k}{k} + \frac12 \log 2\pi \frac{1}{k} + o(1/k)$$
$$ \sum_{k=3}^{n}  \frac{g_{k}}{k} = n  \log n - 2n + o(n) $$
