Finding an asymptotic expansion for $\sum_{k=0}^{n} \frac{1}{1+\frac{k}{n}}$ 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?
 A: Let $ \displaystyle f(x) = \frac{1}{1+\frac{x}{n}}$.
Using the Euler-Maclaurin summation formula, we get
$$ \begin{align} \sum_{k=0}^{n} \frac{1}{1+\frac{k}{n}} &\sim \int_{0}^{n} \frac{1}{1+\frac{x}{n}} \, dx + \frac{f(n)+f(0)}{2} + \sum_{m=1}^{\infty} \frac{B_{2m}}{(2m)!} \left(f^{(2m-1)}(n) -f^{(2m-1)}(0) \right) \\ &= n \ln \left( 1+\frac{x}{n} \right) \Bigg|^{n}_{0} + \frac{\frac{1}{2}+1}{2}  + \frac{1}{6} \left(\frac{1}{2!} \right) \left(-\frac{1}{n} \frac{1}{(1+\frac{x}{n})^{2}} \right) \Bigg|^{n}_{0} \\ &- \frac{1}{30} \frac{1}{4!} \left( - \frac{6}{n^{3}} \frac{1}{(1+ \frac{x}{n})^{4}}\right) \Bigg|^{n}_{0} + \mathcal{O}(n^{-5}) \\ &= n \ln (2) + \frac{3}{4} + \frac{1}{16n} - \frac{1}{128 n^{3}} + \mathcal{O}(n^{-5}) \end{align}$$
For $n=20$, the above approximation is correct to $8$ digits after the decimal point.
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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}
