Showing this sum is $\sim n^{3-\alpha}$.

Question

I want to show that

$$\sum_{i=1}^n i^{2-\alpha} \sim c n^{3-\alpha}$$

for some constant $c$. Here, $\alpha$ is assumed $0 < \alpha < 1$.

I know that

$$\sum_{i=1}^n i^2 \sim c n^3$$

for some constant $c$. So it certainly seems "obvious", but how do I show this?

For starters, we have

$$\sum_{i=1}^n i^{2-\alpha} \geq n^{-\alpha} \sum_{i=1}^n i^2 \sim c n^{3-\alpha}$$

which gives one direction. How can I show the other?

Answer 1

Approximate the sum by the integral $\int_0^n x^{2-\alpha}\,dx$. Let $S_n$ denote this sum, and $I_n$ the integral. Then $I_n\le S_n\le I_{n+1}$. But $I_n= cn^{3-\alpha}$ where $c=1/(3-\alpha)$ and the asymptotic swiftly follows.

Answer 2

Beside Lord Shark the Unknown's simple and good solution, you could also consider that $$\sum_{i=1}^n i^{2-\alpha}= H_n^{(\alpha -2)}$$ where appear generalized harmonic numbers.

Now, using asymptotics $$H_n^{(\alpha -2)}=n^{2-\alpha } \left(\frac{n}{3-\alpha }+\frac{1}{2}+\frac{2-\alpha}{12n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (\alpha -2)$$