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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?

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:

$$psi(z)=-gamma+sum_{n=0}^infty frac{z-1}{(n+1)(n+z)}$$

Yuriy S gave the only possible closed form for the summation.

For large $n$, for sure, you could use series expansions and get from
$$S_n=sum_{i=0}^n frac{1}{a+i,d}=frac 1d left(psi left(n+1+frac{a}{d}right)-psi left(frac{a}{d}right) right)$$
$$S_n=frac{log(n)-psi left(frac{a}{d}right)}{d}+frac{2 a+d}{2 d^2 n}-frac{6 a^2+6 a d+d^2}{12
d^3 n^2}+frac{2 a^3+3 a^2 d+a d^2}{6 d^4
n^3}+Oleft(frac{1}{n^4}right)$$

Using $a=pi$, $d=e$ and $n=10$, the exact result would be $approx 1.03107$ while the above series expansion would give $approx 1.03113$.

The Harmonic Series

The partial sum of the standard Harmonic Series is given by
$$
H_n=sum_{k=1}^nfrac1ktag1
$$
This can be extended to a function that is analytic except at the negative integers
$$
H(z)=sum_{k=1}^inftyleft(frac1k-frac1{k+z}right)tag2
$$
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}^nfrac1{a+(k-1)d}
&=frac1dsum_{k=1}^nfrac1{frac ad+k-1}\
&=frac1dsum_{k=1}^inftyleft(frac1{frac ad+k-1}-frac1{frac{a-d}d+k-1+n}right)\[3pt]
&=frac1dleft(H!left(n+frac ad-1right)-H!left(frac ad-1right)right)tag4
end{align}
$$
using the extension in $(2)$.

Formula $(2)$ from this answer allows us to compute $H!left(frac ad-1right)$ for rational $a$ and $d$, while formula $(3)$ above allows us to approximate $H!left(n+frac ad-1right)$ for large $n$.