Asymptotic expansion of $\sum_{k=0}^n \frac{\ln(k+x)}{(k+x)}$ at $n \to \infty$ Can someone help me get an asymptotic expansion for $$\sum_{k=0}^n \frac{\ln(k+x)}{(k+x)}$$ at $n=\infty$, where $x$ is fixed, I need it with accuracy up to like $O(n^{-3})$, I expect there to be some generalized stieltjes constants in the expansion, but other then that I think it should be just elementary functions. I would appreciate any help.
 A: Let's use Euler Maclaurin as proposed by vonbrand :
$$\sum_{k=0}^n f(k) \sim \int_0^n f(x) dx +\frac {f(0)+f(n)}2+\sum_{k=1}^\infty \frac{B_{2k}}{(2k)!} \bigl(f^{(2k-1)}(n)-f^{(2k-1)}(0)\bigr)$$
with $f(k):=\frac{\ln(k+x)}{(k+x)}$ then 
$$S_n:=\sum_{k=0}^n f(k) \sim \frac 12\left(\ln(n+x)^2-\ln(x)^2+\frac{\ln(x)}{x}+\frac{\ln(n+x)}{n+x}\right)+\sum_{k>0} \frac{B_{2k}}{(2k)!} \bigl(f^{(2k-1)}(n)-f^{(2k-1)}(0)\bigr)$$
with $\ \displaystyle f^{(k)}(n)=(-1)^k\,k!\frac{\ln(n+x)-H_k}{(n+x)^{k+1}}$
$\displaystyle H_k=\sum_{i=1}^k \frac 1i$ the k-th harmonic number
and $\ B_k\ $ a Bernoulli number
Let's rewrite a little the (divergent!) series at the right :
\begin{align}
R_n:&=\sum_{k>0} \frac{B_{2k}}{(2k)!} f^{(2k-1)}(n)\\
R_n:&=\sum_{k>0} \frac{B_{2k}}{(2k)!} (2k-1)!\frac{H_{2k-1}-\ln(n+x)}{(n+x)^{2k}}\\
R_n:&=\sum_{k>0} \frac{B_{2k}}{2k} \frac{H_{2k-1}}{(n+x)^{2k}}-\ln(n+x)\sum_{k>0} \frac{B_{2k}}{2k} \frac 1{(n+x)^{2k}}\\
\end{align}
The $R_n$ series is divergent (as it occurs often with Maclaurin series) and you'll have to interrupt it after some terms. I would conjecture that the maximal number of terms (before decrease of precision) is around $\pi\,x$ for all values of $n \gg 1$.
This means in practice that the precision will be low for small $x$ (in practice your absolute error is bounded by $e^{-2\pi x}$ that may be obtained using nearly $\pi x$ terms).
If you want more precision using the Euler Maclaurin method you will have to compute some terms of the series first (say $m$) and use E-Ml to compute the remainder $\sum_{k=m}^n f(k)$ (i.e. replace $0$ by $m$ in the previous study).

In your question you hoped for an answer in terms of Stieltjes constants. In fact a much more direct answer (ok an equivalence!) is simply :
$$\sum_{k=0}^n \frac{\ln(k+x)}{(k+x)}=\gamma_1(x)-\gamma_1(x+n+1)$$
as you may find in two rather interesting papers of Mark Coffey :


*

*'On representations and differences of Stieltjes coefficients, and other relations' and 

*'Series representations for the Stieltjes constants'

I got too (earlier) an integral representation :
\begin{align}
\tag{1}S_n(x):&=\sum_{k=0}^n \frac{\ln(k+x)}{(k+x)}\\
&=\lim_{p\to -1}\frac {\partial}{\partial p}\sum_{k=0}^n (k+x)^p\\
&=\lim_{p\to -1}\frac {\partial}{\partial p}\left(\zeta(-p,n+x+1)-\zeta(-p,x)\right)\\
\tag{2}&=-\lim_{p\to 1}\frac {\partial}{\partial p}\left(\zeta(p,n+x+1)-\zeta(p,x)\right)\\
\end{align}
(with $\zeta(p,n)$ the Hurwitz zeta function)
This allows to rewrite your problem in integral form (using $(5)$ from the link) :
$$\zeta(p,x)=\frac 1{\Gamma(p)}\int_0^\infty \frac{t^{p-1}\ dt}{e^{xt}(1-e^{-t})}$$
so that :
\begin{align}
\frac {\partial}{\partial p}\zeta(p,x)&=\frac {\partial}{\partial p}\left(\frac 1{\Gamma(p)}\right)\int_0^\infty \frac{t^{p-1}\ dt}{e^{xt}(1-e^{-t})}+\frac 1{\Gamma(p)}\int_0^\infty \frac{\ln(t)\,t^{p-1}\ dt}{e^{xt}(1-e^{-t})}\\
&=\frac 1{\Gamma(p)}\int_0^\infty \frac{\left(-\psi(p)+\ln(t)\right)\,t^{p-1}\ dt}{e^{xt}(1-e^{-t})}\\
\end{align}
(since $\frac d{dx}\frac 1{\Gamma(x)}=-\frac{\psi(x)}{\gamma(x)}$ with $\psi$ the digamma function)
The limit as $\,p\to 1$ gives :
$$\lim_{p\to 1}\frac {\partial}{\partial p}\zeta(p,x)=\int_0^\infty \frac{\gamma+\ln(t)}{e^{xt}(1-e^{-t})}dt$$
and (from $(2)$) we get an integral for $S_n$ :
$$S_n(x)=\int_0^\infty \left(1-\frac 1{e^{(n+1)t}}\right)\frac{\gamma+\ln(t)}{e^{xt}(1-e^{-t})}dt$$
(I didn't reverify this last part...)
