Harmonic series partial sum According to Beyer (1987) the following progression is called harmonic series: $$\frac {1} {a_{1}},\frac {1} {a_{1}+d},  \frac {1} {a_{1}+2d},...$$
How it can be calculated the partial sum of the above mentioned sequence? 
 A: The sum of terms for $n=0,1,...,m$ is $$\sum_{n=0}^m \frac{1}{d(a_1/d+n)} =\frac{1}{d} \left(\psi \left(m+1+\frac{a_1}{d}\right) -\psi \left(\frac{a_1}{d}\right) \right)$$
Where $\psi$ is the Digamma function.
Use the definition:
$$\psi(z)=-\gamma+\sum_{n=0}^\infty \frac{z-1}{(n+1)(n+z)}$$
A: Yuriy S gave the only possible closed form for the summation.
For large $n$, for sure, you could use series expansions and get from
$$S_n=\sum_{i=0}^n \frac{1}{a+i\,d}=\frac 1d \left(\psi \left(n+1+\frac{a}{d}\right)-\psi \left(\frac{a}{d}\right) \right)$$
$$S_n=\frac{\log(n)-\psi \left(\frac{a}{d}\right)}{d}+\frac{2 a+d}{2 d^2 n}-\frac{6 a^2+6 a d+d^2}{12
   d^3 n^2}+\frac{2 a^3+3 a^2 d+a d^2}{6 d^4
   n^3}+O\left(\frac{1}{n^4}\right)$$
Using $a=\pi$, $d=e$ and $n=10$, the exact result would be $\approx 1.03107$ while the above series expansion would give $\approx 1.03113$.
A: The Harmonic Series
The partial sum of the standard Harmonic Series is given by
$$
H_n=\sum_{k=1}^n\frac1k\tag1
$$
This can be extended to a function that is analytic except at the negative integers
$$
H(z)=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+z}\right)\tag2
$$
The Euler-Maclaurin Sum Formula gives the asymptotic expansion
$$
H_n=\log(n)+\gamma+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}-\frac1{252n^6}+\frac1{240n^8}-\frac1{132n^{10}}+O\!\left(\frac1{n^{12}}\right)\tag3
$$
where $\gamma=0.57721566490153286060651209$ is the Euler-Mascheroni Constant.

The Series in the Question
$$
\begin{align}
\sum_{k=1}^n\frac1{a+(k-1)d}
&=\frac1d\sum_{k=1}^n\frac1{\frac ad+k-1}\\
&=\frac1d\sum_{k=1}^\infty\left(\frac1{\frac ad+k-1}-\frac1{\frac{a-d}d+k-1+n}\right)\\[3pt]
&=\frac1d\left(H\!\left(n+\frac ad-1\right)-H\!\left(\frac ad-1\right)\right)\tag4
\end{align}
$$
using the extension in $(2)$.
Formula $(2)$ from this answer allows us to compute $H\!\left(\frac ad-1\right)$ for rational $a$ and $d$, while formula $(3)$ above allows us to approximate $H\!\left(n+\frac ad-1\right)$ for large $n$.
