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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.

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Use the Euler-Maclaurin formula. The main problem will be to determine the constant (in your case dependent on $x$, an extra complication). –  vonbrand Mar 30 '13 at 4:43
    
I'm sorry, let me delete that. –  Vincent Tjeng Mar 30 '13 at 5:02
    
Related: math.stackexchange.com/questions/13300/… –  Aryabhata Mar 30 '13 at 5:04
    
@vonbrand: Will you be willing to add an answer? –  Aryabhata Mar 30 '13 at 5:15
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1 Answer

up vote 3 down vote accepted

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 :


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...)

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I appreciate all the time you spent writing this, I also new there should be a closed form in terms of stieltjes constants, if one allows integer values in them, but to me this is kind of like not reducing values of the gamma function to there fractional part, or like saying there is a closed form for harmonic numbers, and just re-writing the nth harmonic number in terms of the digamma function. Nice answer anyway, I found what I was looking for by shifting through your articles and a couple others I found. –  Ethan Mar 31 '13 at 3:00
    
Glad it helped @Ethan. Yes there is much rewriting in these subjects (zeta, polylogs...) but sometimes one of these expressions allows fine approximations or better links with other interesting subjects. Could you write somewhere the asymptotic expression you obtained? Thanks, –  Raymond Manzoni Mar 31 '13 at 9:41
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