# 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 $$\int_1^n \frac{dx}{x^\alpha}.$$
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 $\frac{n^{1-\alpha}}{1-\alpha}$.