So I have for simpler notation, for a fixed infinitely large natural number $n$ and all finite natural numbers $m,z$, then $L(n+m)=\log(n+m)!$, and this is equal to $$L(n+m)=\log n!+\sum_{k=1}^m\log(n+k)$$
But for sake of clarity, if I include additional finite natural numbers, does the equation become $$L(n+m+z)=\log n!+\sum_{k=1}^z\sum_{k=1}^m\log(n+k)$$ Do I sum over each finite natural number, and so what is the general case?