divergence rate for a sum involving log I ma trying to see how $$S_m=\sum_{h=1}^{m-1}\frac{m-h}{h}\log h$$
Numerically I get the following graph, after trying different simple "guessing" for the rate I think that the rate is not a simple function. 

The red line is $m (\log m)^{1.6} $ and the blue if the sum ...
I appreciate any hints.
 A: Using the Euler-Maclaurin Sum Formula, we get
$$
\sum_{h=1}^{m-1}\frac{m-h}{h}\log(h)=\small m\left(\frac{\log(m)^2}2-\log(m)+1+\gamma_1\right)-\frac{\log(2\pi)}2-\frac{\log(m)}{12m}+O\left(\frac{\log(m)}{m^3}\right)
$$
where $\gamma_1=-0.07281584548367672486$ is one of the Stieltjes Constants.
So using the asymptotic approximation
$$
m\left(\frac{\log(m)^2}2-\log(m)+1+\gamma_1\right)-\frac{\log(2\pi)}2
$$
the error gets smaller as $m$ gets larger.

Computation of $\boldsymbol{\gamma_1}$
$$
\gamma_1=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{\log(k)}k-\frac{\log(n)^2}2\right)
$$
The Euler-Maclaurin Sum Formula says that
$$
\begin{align}
\small\sum_{k=1}^n\frac{\log(k)}k
&\small=\frac{\log(n)^2}2+\gamma_1+\frac{\log(n)}{2n}-\frac{\log(n)-1}{12n^2}+\frac{6\log(n)-11}{720n^4}-\frac{60\log(n)-137}{15120n^6}\\
&\small+\frac{140\log(n)-363}{33600n^8}-\frac{1386\log(n)-7129}{332640n^{10}}+O\left(\frac{\log(n)}{n^{12}}\right)
\end{align}
$$
Setting $n=100$, we get $\gamma_1$ to over $20$ places.
