How to find the the sum of the finite series $ \sum_{n=1}^{N-1} \frac{1}{n-k+1}$ ?
If I expand the series then, we get
$ \sum_{n=1}^{N-1} \frac{1}{n-k+1}=\frac{1}{2-k}+\frac{1}{3-k}+\cdots+\frac{1}{N-k}$.
How to sum it.
Here $N,k \in \mathbb{N} \cup \{0\}$.
Is there any short or closed form of it in terms of probably $N, k$?
There is no closed form in terms of elementary functions. In terms of the harmonic numbers, defined by
$$H_m=\sum_{n=1}^m\frac1n$$
we have
$$H_{N-k}-H_{1-k}=\sum_{n=1}^{N-1}\frac1{n-k+1}$$
supposing that $k$ is a negative integer. For non-integer $k$, we can use the above as generalized harmonic numbers defined by
$$H_x=\sum_{n=1}^\infty \left(\frac1n-\frac1{n+x}\right)$$
or equivalent, such as the digamma function, Hurwitz zeta function, etc.