What are the $p$ and $q$ for the sum in order to converge

Question

For what values of $p$ and $q$ does the following sum converge?$$\sum_{n=2}^{\infty}\frac{1}{ n^p-n^q}\quad(0<q<p)$$

I really can't wrap my head around this. Can anyone provide a hint and not the solution?

Answer 1

Presumably, you already know how to choose $p$ such that $\sum_{n}\frac{1}{n^p}$ converges or diverges.

So, let's think asymptotically: when $n\to\infty$, using the assumption that $0<q<p$, what is the limiting behavior of $\frac{1}{n^p-n^q}$? If you were going to write its asymptotic behavior as $\frac{1}{n^x}$ for some $x$, what $x$ would that be?

From here, you can use something like the Limit Comparison test to establish that the behavior of the given series is the same as the limit of the sum of the "approximate" $p$-series you came up with.

Answer 2

For $p>1$ compare $n^p-n^q=n^p(1-\frac {1}{n^{p-q}})$ to $\frac {1}{2}n^p$ for large $n.$

For $p\leq 1$ compare $n^p-n^q$ to $n^p.$