Let $0 < \alpha < 1$. Can somebody please explain why $$\sum_{i=1}^n \frac{1}{i^\alpha} \sim n^{1-\alpha}$$ holds?
Asymptotic growth of $\sum_{i=1}^n \frac{1}{i^\alpha}$?
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$\begingroup$
sequences-and-series
asymptotics
2 Answers
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Compare with the integral $$\int_1^n \frac{dx}{x^\alpha}.$$
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0This falls short of showing why this is the _dominant_ contribution. Or is it obvious? – 2012-10-07
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0@Sasha: One actually wants inequalities that use two closely related integrals. But I thought something should be left for the OP to do. – 2012-10-07
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0Thanks! 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](http://math.stackexchange.com/questions/208152/estimate-a-sum-of-products) of mine? – 2012-10-07
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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 $\frac{n^{1-\alpha}}{1-\alpha}$.