Proving An Asymptotic Relation Needed to Answer a Probability Problem

Question

I am working on understanding problem 3.4.13 from Durrett's Probability Theory and Examples. I found a solution online that relies on the following relationship: $$ \sum\limits_{j=1}^n j^{2-\beta}\sim \frac{n^{3-\beta}}{3-\beta}. $$

I have no idea how you show something like this.

Could anyone give some help?

Thank you.

Answer 1

$\sum_{j=1}^n j^{2-\beta}$ is a Riemann sum for $\int_0^n x^{2-\beta} dx$, which is equal to the right side you have given. (Specifically it is the right hand sum with $h=1$.) You can then use the error estimate for the rectangle rule to prove that the difference is of lower order than $n^{3-\beta}$.

Answer 2

We can approximate sums with integrals. If $f$ is an increasing function, then $$ \int_{s=a-1}^{b} f(s)\ ds \le \sum_{i=a}^{b} f(i) \le \int_{s=a}^{b+1} f(s)\ ds. $$ If $f$ is a decreasing function, then $$ \int_{s=a}^{b+1} f(s)\ ds \le \sum_{i=a}^{b} f(i) \le \int_{s=a-1}^{b} f(s)\ ds. $$ The notation $\sim$ means that the ratio of the left side and the right side tends to $1$ as $n$ goes to infinity.