TITLE: Finding an asymptotic expansion for $\sum_{k=0}^{n} \frac{1}{1+\frac{k}{n}}$
QUESTION [7 upvotes]: It is well known that an asymptotic expansion of the n-th harmonic number is $$H_{n}= \sum_{k=1}^{n} \frac{1}{k} \sim  \ln(n) + \gamma + \frac{1}{2n} -\frac{1}{12n^{2}} + O(n^{-4}).$$
How could we find an asymptotic expansion for the sum $ \displaystyle \sum_{k=0}^n \frac{1}{1+\frac{k}{n}}$ to a similar order?

REPLY [3 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Since the series has a closed expression in terms of the Digamma Functions, it's interesting to check the expansion by means of the asymptotic behavior of those functions:
\begin{align}&\color{#66f}{\large\sum_{k=0}^{n}\frac{1}{1 + k/n}}
=n\sum_{k=0}^{n}\frac{1}{k + n}
=n\sum_{k=0}^{\infty}\pars{\frac{1}{k + n} - \frac{1}{k + 2n + 1}}
\\[5mm]&=n\bracks{\Psi\pars{2n + 1} - \Psi\pars{n}}\tag{1}
\end{align}
where $\ds{\Psi}$ is the Digamma Function and we used
${\bf 6.3.16}$. With
${\bf 6.3.5}$, expression $\pars{1}$ becomes:
\begin{align}&\color{#66f}{\large\sum_{k=0}^{n}\frac{1}{1 + k/n}}
=n\bracks{\Psi\pars{2n} - \Psi\pars{n}} + \half
\end{align}
With the
Digamma Asymptotic Expansion
${\bf 6.3.18}$:
\begin{align}&\color{#66f}{\large\sum_{k=0}^{n}\frac{1}{1 + k/n}}
\sim n\times
\\[5mm]&\braces{\!\!\bracks{%
\ln\pars{2n} - \frac{1}{4n} - \frac{1}{48n^{2}} + \frac{1}{1920n^{4}}
-\frac{1}{16128n^{6}}}\!\!
-\!\!\bracks{%
\ln\pars{n} - \frac{1}{2n} - \frac{1}{12n^{2}} + \frac{1}{120n^{4}}
-\frac{1}{252n^{6}}}}
\\[5mm]&-\half
\\[1cm]&=n\bracks{\ln\pars{2} + \frac{1}{4n} + \frac{1}{16n^{2}} - \frac{1}{128n^{4}} + \frac{1}{256n^{6}}} + \half
\\[5mm]&=\color{#66f}{\large n\ln\pars{2} + \frac{3}{4}
+ \frac{1}{16n} - \frac{1}{128n^{3}} + \frac{1}{256n^{5}}}
\end{align}