Asymptotic behavior of $\sum\limits_{k=1}^n \frac{1}{k^{\alpha}}$ for $\alpha > \frac{1}{2}$

Question

As the title states, I'm interested in the asymptotic behavior of $$\sum\limits_{k=1}^n \frac{1}{k^{\alpha}} , \alpha > \frac{1}{2}$$ for $n \to \infty $. Any hints/ideas?

Answer 1

$ \forall a\in (0,1)$ By comparison to integrals, $$ \frac{1}{1-a}\Big((n+1)^{1-a}-1\Big)=\int_1^{n+1}\frac{\mathrm{d}x}{x^a}\le\sum_{k=1}^n\frac1{k^a}\le\int_0^n\frac{\mathrm{d}x}{x^a}=\frac{1}{1-a}n^{1-a} $$ So$\sum_{k=1}^n\frac1{k^a}=O(n^{1-a})$

$\sum_{k=1}^n\frac1{k}\sim \ln n$

Answer 2

As shown in this answer, we can use the Euler-Maclaurin Sum Formula to get the asymptotic expansion $$ \sum_{k=1}^nk^{-a}=\zeta(a)+\frac1{1-a}n^{1-a}+\frac12n^{-a}-\frac{a}{12}n^{-1-a}+\frac{a^3+3a^2+2a}{720}n^{-3-a}+O(n^{-5-a}) $$ For $a\gt1$, this shows that $$ \sum_{k=1}^\infty k^{-a}=\zeta(a) $$ where $\zeta$ is the Riemann Zeta Function.

The zeta function has a singularity at $1$, so we handle that case separately: $$ \sum_{k=1}^nk^{-1}=\gamma+\log(n)+\frac12n^{-1}-\frac1{12}n^{-2}+\frac1{120}n^{-4}+O(n^{-6}) $$ where $\gamma$ is the Euler-Mascheroni Constant.

Answer 3

$$\sum_1^n\dfrac1{k^a}~\simeq~\int_1^n\dfrac{dx}{x^a}~=~?$$