Asymptotic equation for $\sum_{k=1}^{n} \frac{\ln(k)^m}{k}$ for $m=1,2,3,4; n\gg0$?

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Is there a known asymptotic equation for $$ S(m)=\sum_{k=1}^{n} \frac{\ln(k)^m}{k} $$

where in particular $m=1,2,3,4$ and it is assumed that $n\gg0$. When looking at the interpolations of 2nd order polynomials, it shows that for $k \in [1,n=819]$

$$ S(1)=-3*10^{-5}n^2+0.0422n+5.8118 $$

$$ S(2)=-1*10^{-4}n^2+0.1964n+12.024 $$

$$ S(3)=-4*10^{-4}n^2+0.9333n+19.718 $$

The logarithmic interpolations are less accurate for something like $a\ln(n)+b$ for some constant $a,b$.

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M. I. Israilov showed that for any $m\ge 0$, $n\ge 2$ and $N\ge 1$, \begin{align*} \sum\limits_{k = 1}^n {\frac{{\ln ^m k}}{k}} = \frac{{\ln ^{m + 1} n}}{{m + 1}} + \frac{{\ln ^m n}}{{2n}} + \gamma _m & + \sum\limits_{k = 1}^{N - 1} {\frac{{B_{2k} }}{{(2k)!}}\left[ {\frac{{{\rm d}^{2k - 1} }}{{{\rm d}x^{2k - 1} }}\frac{{\ln ^m x}}{x}} \right]_{x = n} } \\ & + \theta _N (m,n)\frac{{B_{2N} }}{{(2N)!}}\left[ {\frac{{{\rm d}^{2N - 1} }}{{{\rm d}x^{2N - 1} }}\frac{{\ln ^m x}}{x}} \right]_{x = n} \end{align*} with a suitable $0<\theta _N (m,n)<1$. Here, the $\gamma_m$ are the Stieltjes constants and the $B_k$ denote the Bernoulli numbers.