What is the solution for the following sum?
$\sum\_{i=1}^n i^{-\alpha}$ for $-2\leq \alpha\leq 0$. 


For example if we have $\alpha =-1 $ we know it is the harmonic series thus $=\log n$ and for $\alpha=-2$ it is $\Theta(1)$. What if $\alpha$ ranges? Is there any more general formula?


How about if $\alpha>1$? Can we say it is $\Theta(n^{1+\alpha})$? I can easily be verified by Gauss trick that $O(n^{1+\alpha})$ holds. How about the lower bound?



Sum of series $\sum_{i=1}^n i^{-\alpha}$
$$\sum\_{i=1}^n i^{-\alpha}=H\_{n,\alpha}$$
the [generalized Harmonic number.](https://en.wikipedia.org/wiki/Harmonic_number#Generalized_harmonic_numbers) For $\alpha>1$ one has the limit
$$\lim\_{n\rightarrow\infty}H\_{n,\alpha}=\zeta(\alpha),$$
the [Riemann zeta function.](https://en.wikipedia.org/wiki/Riemann_zeta_function) The large-$n$ asymptotics is
$$H\_{n,\alpha}=\zeta(\alpha)-\frac{1}{n^\alpha}\sum\_{k=-1}^\infty\frac{B\_{k+1}}{(k+1)!}\frac{(\alpha)\_k}{n^k},$$
with $B\_{k+1}$ Bernoulli numbers and $(\alpha)\_k$ rising factorials. (See this [MSE posting.)](https://math.stackexchange.com/questions/1405208/asymptotic-of-sum-r-1k-frac1r3-2-generalized-harmonic-number)


