Infinite Limit of an Infinite Power Series I've been stuck on this question for some time now:

For $\alpha, \beta > -1$, what is the value of  $\lim \limits_{n \to \infty} n^{\beta-\alpha} \frac{1^\alpha + 2^\alpha + \dots + n^\alpha}{1^\beta + 2^\beta + \dots + n^\beta}$?

I've never tackled such limits before — most of my dealings with infinite limits have been with terms that can be easily reduced to the form $\frac{1}{n}$, and then simply replacing them with zero, usually yielding a fraction behind. Clearly this is not of that form. Furthermore the exact values of $\alpha$ and $\beta$ have not been specified either.
I thought of using L'hopital's rule, but doing so left behind an extremely messy expression which left me quite sure that that was not the method to go.
I would appreciate a hint, but in general how to go about solving such limits?
 A: $$L = \lim_{n \to \infty} \frac{n^\beta}{n^\alpha} \times \frac{\sum_{r=0}^n r^\alpha}{\sum_{r=0}^n r^\beta}\\
= \lim_{n \to \infty} \frac{\sum_{r=0}^n (\frac rn)^\alpha}{\sum_{r=0}^n (\frac rn)^\beta}$$
The summations can now be evaluated as Riemann sums by multiplying numberator and denominator by $\frac 1n$
$$L = \frac{\int_0^1 x^\alpha dx}{\int_0^1 x^\beta dx}\\
L = \frac{\beta+1}{\alpha +1}$$
A: Without Riemann sums.
$$\sum_{r=1}^n r^k=H_n^{(-k)}$$ where appear the generalized harmonic numbers. Their asymptotics is given by
$$ H_n^{(-k)}=n^k \left(\frac{n}{k+1}+\frac{1}{2}+\frac{k}{12
   n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-k)$$
$$ n^{\beta-\alpha} \frac{H_n^{(-\alpha )}}{H_n^{(-\beta )}}= n^{\beta-\alpha}\frac{n^{\alpha } \left(\frac{n}{\alpha +1}+\frac{1}{2}+\frac{\alpha }{12
   n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-\alpha ) } {n^{\beta } \left(\frac{n}{\beta +1}+\frac{1}{2}+\frac{\beta }{12
   n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-\beta ) }$$ which is
$$\sim\frac{\beta +1}{\alpha +1}+\frac{(\beta +1) (\alpha -\beta )}{2 (\alpha +1) n}+O\left(\frac{1}{n^2}\right)$$
