Vtyjkyuk

Is there a similar formula like the Stirling one on the sum over $ln(n)$ (take logarithms on its factorial representation),
$$
sum_n=1^N ln(n) = Ncdot ln(N)−N+fracln(N)2+fracln(2π)2+Oleft(fracln(N)Nright),
$$
but for the following sum?

$$sum_n=1^N H_n cdot ln(n)$$

I already advanced on getting good approximation on asymptotics using Euler-McLaurin approximation on summation terms till $O(ln(N))$ order. But further advance is becoming hard for me in $O(1)$ term. This is often referred as the Ramanujan summation of this series.

The constant term is difficult to construct. I'll present the terms to $O(1/n)$ which includes the constant. This problem was asked in MSE 2891159, in which case I answered one of three; this problem is that problem's $sigma_c(n).$ The problem can be setup exactly as that previous problem, by differentiating with respect to $s$ the following, and taking limits when needed:
$$ sum_k=1^n k^-s=zeta(s)-frac1(s-1)n^s-1+frac12n^s-
frac112fracsn^s+1 +.... $$
Higher terms are neglected because what is present is sufficient to get to $O(1/n)$ terms. The formula follows from the Euler-McLaurin summation. As in the other problem, break the sum you want into primitives consisting of sums over $logk/k^m$ or $log^2k/k^m.$ That is,
$$sum_k=1^n H_k logk = sum_k=1^n Big(gamma + logk + frac12k -frac112k^2 Big) logk$$
where the asymptotic formula for $H_k = gamma + psi(k+1)$ has been used and a sufficient number of terms have been taken, $except$ for those need to derive the constant term (we'll come back to that later). Letting $L=logn$ we have the following:
$$ v_0=sum_k=1^n log^2k=nbig(L^2+2(1-L)) + fracL^22+fracL6n+fracgamma^22-fracpi^224-frac12log^2(2pi)+gamma_1$$
$$ quad v_1=sum_k=1^n logk=n(L-1)+fracL2-zeta'(0)+frac112n$$
$$v_2= sum_k=1^n fraclogkk =fracL^22+gamma_1+fracL2n$$
$$ v_3=sum_k=1^n fraclogkk^2 =-zeta'(2)-fracL+1n$$
Let $tildev_k=(v_k$ with constant term set to 0). Then
$$sigma_c(n)=sum_k=1^n H_k logk =tildev_0+gamma, tildev_1 + frac12 tildev_2-frac112 tildev_3 + C $$
$$=nbig(L^2+(gamma-2)(L-1)big) + frac3L^24+gammafracL2+frac12n(L+fracgamma+16) + C $$
where $C$ is the unknown constant term. A correct way to determine C is simply by
$$ C=lim_n to infty Big(sum_k=1^n H_k logk - Big(nbig(L^2+(gamma-2)(L-1)big) + frac3L^24+gammafracL2 Big), Big)$$
Proceed further only if you are comfortable with formal mathematics. Now I'm going to find $C$ in a different manner. It is well-known that
$$ gamma = lim_n to inftyBig( sum_k=1^nfrac1k - logn Big) =
int_0^infty Big(frac1e^x-1 - frace^-xx Big)dx.$$ The first expression is similar in spirit to how we defined $C,$ but the second equation is useful in that many digits of $gamma$ can be extracted quite easily with your favorite numerical integration routines.

In derving $v_0$ through $v_3$ I stopped because this is sufficient to get terms to $O(1/n).$ If we kept going, there would be a $v_5 = sum_k=1^n logk/k^4= -zeta'(4) + o(1/n)$, etc. The natural thing to do is add all the $-zeta'(2n)$ terms up with the appropriate weights to get a new constant that, when added to the ones in the $v_k$, constitute $C.$ This was done in MSE 2891159 and very fortunately the series converged. No such luck here. The weights are from the asymptotic expansion of the harmonic numbers,
$$H_k = gamma + logk + frac12k-sum_m=1^infty fracB_2m2m k^-2m $$ so the constant we want to assign meaning to is
$$ kappa ,,dot= sum_m=1^infty fracB_2m2m zeta'(2m) $$
where the dotted equals means 'representation' instead of 'equals.' The idea is akin to saying $$sum_k=1^infty k ,, dot= -1/12$$ which can be given rigor in terms of zeta regularization. First differentiate the well-known Euler-like integral for the zeta function to find
$$zeta'(2s) =frac1Gamma(2s) int_0^infty fract^2s-1e^t-1 Big( logt - psi(2s) Big) .$$
Insert into the definition for $kappa$ and interchange $int$ and $sum$

$$ (K) quad kappa ,,dot= int_0^infty fracdt/te^t-1sum_m=1^infty fracB_2m(2m)! t^2m big( logt - psi(2m) big). $$ The inner sum converges (its been Borel transformed), we just need to find an expression that's not a power series. We need some formulas. It is well-known that, and it will be used repeatedly,
$$sum_m=1^infty fracB_mm! t^m = fracte^t-1 -1, quad sum_m=1^infty fracB_2m(2m)! t^2m= fracte^t-1 -1+t/2 $$
An integration of the previous equation leads to
$$sum_m=1^inftyfracB_2m(2m)(2m)! t^2m= -logbig(fracte^t-1big) -t/2 $$ Let's begin on the 'psi' term:
$$ Psi(t):=sum_m=1^infty fracB_2m(2m)! t^2m ,psi(2m)=
sum_m=1^infty fracB_2m(2m)! t^2m ,big(psi(2m)+frac12m - frac12m big)=$$
$$= sum_m=1^infty fracB_2m(2m)! t^2m ,psi(2m+1)+Big(t/2+logbig(fracte^t-1big) Big) =$$
$$= sum_m=1^infty fracB_mm! t^m ,psi(m+1)+
Big( frac12(1-gamma)t Big) +
Big(t/2+logbig(fracte^t-1big) Big) .$$
Again use $H_m = gamma + psi(m+1)$ to find
$$Psi(t)=sum_m=1^infty fracB_mm! t^m ,H_m -
gammaBig(fracte^t-1 - 1Big)+ big(1-fracgamma2big)t + logbig(fracte^t-1big) $$
The harmonic number is used because there is the integral relation
$$H_m = -m int_0^1 dx x^m-1 log(1-x) $$ Insert this, switch $int$ and $sum$, sum up the series in closed form, and $finally$
$$Psi(t)=-int_0^t fraclog(1-u/t)e^u-1Big(1-fracu,e^ue^u-1Big),du-
gammaBig(fracte^t-1 - 1Big)+ big(1-fracgamma2big)t + logbig(fracte^t-1big) $$
In eq. $(K)$ the sum before the $logt$ is one of the known formulas. Therefore a have a (double) integral relationship that is perfectly well-behave for $kappa$
$$kappa=int_0^infty fracdt/te^t-1Big(, logtbig(fracte^t-1-1+t/2big) - Psi(t)Big) = -0.077596...$$
When using $kappa$ the answer can be stated as
$$ C=kappa + fracgamma^22-fracpi^224+frac12gamma , log(2pi) - frac12 log^2(2pi) + frac32 gamma_1 $$
Six digits of agreement are obtained by using the asymptotic formula and comparing to the brute force summation for $n=$ 20.

This is not a solution by an extended comment.

It might be interesting to see to which extent the final exact result for the divergent sum (notice that $kappa_c$ is defined here as a positive quantity)

$$kappa_c dot= - sum_n=1^infty fracB(2n)2nzeta'(2n)$$

obtained ingeniously by skbmoore is approximated by a finite number n of summands.

The graph below shows that there in the range $3lt nlt 10$ we have reasonable agreement. Outside that region the divergence hits and spoils the result.

Such plots might also in other cases give a feeling of the number of terms to be taken into account and of the value of the sum. Exact results are of course a different topic.