Given that:
$\mbox{A well-known Euler's sum is}\displaystyle\quad% \sum_{k = 1}^{\infty}\left[{1\over k} - \ln\left(k + 1 \over k\right)\right] = \gamma\label{1}\tag{1} $ The generalised of \eqref{1} is:
$$ \sum_{k = 1}^{\infty}\left[{1 \over kx} + {1 \over kx - 1} + {1 \over kx - 2} + \cdots + {1 \over kx - x + 1} - \ln\left(k + 1\over k\right)\right] = \gamma + \ln\left(x\right)\label{2}\tag{2} $$
Where $\gamma$ is Euler's constant and $x \geq 1$.
Is the sum of \eqref{2} correct ?
I'm assuming that $x$ is a positive integer. Note that $$\sum_{m=0}^{x-1}\frac{1}{kx-m}=H_{kx}-H_{x\left(k-1\right)}$$ where $H_{l}$ is the $l$-th harmonic number and since $$\lim_{n\rightarrow\infty}\left(H_{n}-\log\left(n\right)\right)=\gamma$$ we have $$\lim_{N\rightarrow\infty}\left(\sum_{k=1}^{N}\left(\sum_{m=0}^{x-1}\frac{1}{kx-m}-\log\left(\frac{k+1}{k}\right)\right)\right)$$ $$=\lim_{N\rightarrow\infty}\left(\sum_{k=1}^{N}\left(H_{kx}-H_{x\left(k-1\right)}-\log\left(k+1\right)+\log\left(k\right)\right)\right)$$ $$=\lim_{N\rightarrow\infty}\left(H_{Nx}-\log\left(N\right)\right)=\lim_{N\rightarrow\infty}\left(H_{Nx}-\log\left(Nx\right)+\log\left(x\right)\right)=\color{red}{\gamma+\log\left(x\right)}.$$