# Asymptotic growth of $\sum_{i=1}^n \frac{1}{i^\alpha}$?

Let $0 < \alpha < 1$. Can somebody please explain why $$\sum_{i=1}^n \frac{1}{i^\alpha} \sim n^{1-\alpha}$$ holds?

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Compare with the integral $$\int_1^n \frac{dx}{x^\alpha}.$$

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This falls short of showing why this is the dominant contribution. Or is it obvious? –  Sasha Oct 7 '12 at 17:12
@Sasha: One actually wants inequalities that use two closely related integrals. But I thought something should be left for the OP to do. –  André Nicolas Oct 7 '12 at 17:21
Thanks! I guess that is what Davide has done in his answer. While you're at it, could you please have a look at a somewhat related question of mine? –  Haderlump Oct 7 '12 at 17:28

We have for $x\in [k,k+1)$ that $$(k+1)^{—\alpha}\leq x^{-\alpha}\leq k^{—\alpha},$$ and integrating this we get $$(k+1)^{—\alpha}\leq \frac 1{1-\alpha}((k+1)^{1-\alpha}-k^{\alpha})\leq k^{—\alpha}.$$ We get after summing and having used $(n+1)^{1-\alpha}-1\sim n^{1-\alpha}$, the equivalent $\frac{n^{1-\alpha}}{1-\alpha}$.

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Thank you, Davide –  Haderlump Oct 7 '12 at 17:28