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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?

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  • $\begingroup$ Did you mean: $\sum_{i=-n}^{-1}\,(-i)^{-\alpha}\ \mbox{for}\ -2\le\alpha\le 0\ $ ? $\endgroup$
    – Wlod AA
    Commented Jul 5, 2017 at 21:30

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$$\sum_{i=1}^n i^{-\alpha}=H_{n,\alpha}$$ the generalized Harmonic number. For $\alpha>1$ one has the limit $$\lim_{n\rightarrow\infty}H_{n,\alpha}=\zeta(\alpha),$$ the 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.)

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