Asymptotics behaviour of the sum

Question

Let $S(n) = \sum_{k = n}^\infty \frac{ln^q k}{k^p}$, $p > 1$

How to determine asymptotic behaviour of the sum $S(n)$, ($n \rightarrow \infty$).

The use of integrals doesn't solve the problem.

Also I used Stolz–Cesàro theorem to solve the problem, but I can't.

What is the first step of solving?

$ S(n) $ ~ ??

Maybe I need to know sum for k from 1 to infinity and the sum for k from 1 to n. And next subtract one from another. — J.Exactor, Oct 14 '15 at 17:16
Oh sorry. I misread the index. I thought you were summing from 1.. — Vim, Oct 14 '15 at 17:18

score 2 · Answer 1 · answered Oct 14 '15 at 17:34

2

Why do you say integrals are of no use? $$ \int_{n+1}^\infty\frac{(\ln x)^q}{x^p}\,dx\le\sum_{k=n}^\infty\frac{(\ln k)^q}{k^p}\le\int_{n}^\infty\frac{(\ln x)^q}{x^p}\,dx. $$ From here you get (integrating by parts) $$ S(n)\sim\frac{(\ln n)^q}{(p-1)\,n^{p-1}}. $$

answered Oct 14 '15 at 17:34

Julián Aguirre

76,354

Apparently I was afraid such an integral and stopped. – J.Exactor Oct 14 '15 at 17:42

Asymptotics behaviour of the sum

1 Answers1