Is there any way to find the following limit
$$R(n,m)=\lim_{N\to\infty}\frac{H_{nN,m}}{H_{N,m}}$$
which involves harmonic numbers (generalized if $m\neq 1$)
$$H_{N,m}=\sum_{k=1}^N k^{-m}\qquad ?$$
I am more specifically looking for a convenient way to compute it numerically for $m<1$ (if it converges to something else than 1 of course).
From numerical experiment on Mathematica for $m \leq 1$, I can guess $$R(n,m)=n^{1-m} \quad .$$
Suppose first that $0\le m<1$. Then, using the inequality $k^{-m}\ge\int_k^{k+1} x^{-m}\,dx$ for $k>0$, we have $$H_{N,m}\ge\int_1^{N+1}x^{-m}\,dx=\frac{(N+1)^{1-m}-1}{1-m}\sim\frac{N^{1-m}}{1-m}\tag{1}$$ (as $N\to\infty$). Similarly, using the inequality $k^{-m}\le\int_{k-1}^k x^{-m}\,dx$ for $k>1$, we have $$H_{N,m}\le1+\int_1^{N}x^{-m}\,dx=1+\frac{N^{1-m}-1}{1-m}\sim\frac{N^{1-m}}{1-m}.\tag{2}$$ So, here $H_{N,m}\sim\frac{N^{1-m}}{1-m}$. Similarly, $H_{nN,m}\sim\frac{(nN)^{1-m}}{1-m}$. Thus, we confirm that $H_{nN,m}/H_{N,m}\sim n^{1-m}$, if $0\le m<1$.
The case $m<0$ is similar, now with the inequalities in (1) and (2) going in the opposite direction.
Finally, for $m=1$ we have $H_{N,m}\sim\ln N$, whence $H_{nN,m}/H_{N,m}\to1= n^{1-m}$.
