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 
$$R(n,m)=n^{1-m} \quad .$$ 



Limit of a ratio of harmonic numbers?
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 $H\_{nN,m}/H\_{N,m}\to1= n^{1-m}$.


