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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.

Haoran Chen
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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, $$\begin{aligned} \sum_{k=2}^n (H_n-H_{k-1}) \log k &= \sum_{k=2}^n \big(\log n - \log k + O(\tfrac 1k)\big)\log k \\ &= \log n \sum_{k=2}^n \log k - \sum_{k=2}^n \log^2 k + \sum_{k=2}^n O(\tfrac 1k) \log k \\ &= \log n\big(n\log n - n + O(\log(2+n)\big) \\ &- \big(n\log^2n - 2n \log n + 2n + O(\log^2(n+2)\big)\\ &+O\Big(\sum_{k=2}^n \frac{\log k}k \Big)\\ &= n\log n - 2n + O(\log^2(2+n)) \end{aligned}$$

D.R.
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  • Good job! Do you use $\log(n+2)$ rather than $\log(n)$ just to avoid issues when $n$ is small? – Haoran Chen Dec 03 '22 at 10:05
  • @HaoranChen yes, exactly. It is "standard practice", as you see in the Tao notes. You can also restrict $n\geq 2$ for example, in which case $\log(n)$ and $\log(2+n)$ are bounded by constant multiples of each other, hence can be substituted for each other in the big-O notation. – D.R. Dec 03 '22 at 19:57
  • @HaoranChen For your interest, a somewhat more precise asymptotics is $$ \left( n + \frac{1 + \log (2\pi )}{2} \right)\log n - 2n + \frac{\log ^2 n}{4} + C + o(1), $$ with some constant $C=1.547\ldots$. – Gary Dec 05 '22 at 03:07
  • @Gary maybe submit this as another answer along with how you derived it? – D.R. Dec 05 '22 at 04:13
  • @D.R. It is essentially the same derivation as yours except with more precise asymptotics here and there. – Gary Dec 05 '22 at 23:16
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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,

$$ g_{k+1} = k \log k - k +\frac12 \log k + \frac12 \log 2\pi + o(1)$$

$$ \frac{g_{k}}{k} = \log k -1 + \frac32 \frac{\log k}{k} + \frac12 \log 2\pi \frac{1}{k} + o(1/k)$$

$$ \sum_{k=3}^{n} \frac{g_{k}}{k} = n \log n - 2n + o(n) $$

leonbloy
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    When $g_{k+1}$ is replaced by $g_k$, $k$ becomes $k-1$, and I guess the term $\frac{3}{2}\frac{\log k}{k}$ should be $-\frac{1}{2}\frac{\log k}{k}$? It doesn't affect the answer though. – Haoran Chen Dec 03 '22 at 14:15