Convergence of $\sum_{n=2}^\infty \frac{1}{(n(\log n)^p )}$ using the integral test

Question

Prove that $$\sum_{n=2}^\infty \frac{1}{(n(\log n)^p )}$$ converges if and only if $p>1$.

I know that $\sum_{n=2}^\infty f(n)$ converges if and only if $\int_2^\infty f(x)$ converges by the integral test, provided that $f$ is a positive decreasing function on $[2,\infty)$. However, for some values of $p$, $f(x)=\frac{1}{x(\log x)^p}$ is decreasing, as shows the derivative $f(x)=-\frac{1+px}{x^2(\log x)^{p-1}}$. Can we apply the integral test whatsoever or do we need to account for changes? Can we apply the integral test at all?

I have to use the integral test.

@Masacroso the theorem in our book says: let $(a_n){n=1}^\infty$ a sequence such that $a_n=f(n)$ for a decreasing positive function $f:\mathbf{R}\to\mathbf{R}$. Then we have $\sum{n=1}^\infty a_n<\infty \iff \int_1^\infty f(x),dx<\infty$. — Dr. Heinz Doofenshmirtz, Feb 10 '18 at 09:45
@Heinz you are right, sorry. I forget about that. $f$ must be monotone and of one sign. Well if $p>0$ we have that $f$ is decreasing at some point so you can apply the integral test. — , Feb 10 '18 at 09:47
Possible duplicate of On convergence of Bertrand series $\sum\limits_{n=2}^{\infty} \frac{1}{n^{\alpha}\ln^{\beta}(n)}$ where $\alpha, \beta \in \mathbb{R}$ — Guy Fsone, Feb 10 '18 at 10:04

score 3 · Accepted Answer · answered Feb 10 '18 at 09:50

3

For $p>0$, you can apply the integral test since $f$ is decreasing from a certain moment because the first terms of a sum do not change its nature. And I think you know $\sum \frac{1}{n}$ diverges so you can conclude for all $p$.

answered Feb 10 '18 at 09:50

This is what I was looking for. For $p\leqslant 0$, we have that $\frac{1}{n(\log n)^p}>\frac{1}{n}$ and for $p> 0$ we can use the integral test. Thanks! – Dr. Heinz Doofenshmirtz Feb 10 '18 at 09:51

score 2 · Answer 2 · answered Feb 10 '18 at 09:47

2

You can use the Cauchy condensation test

$$ 0 \ \leq\ \sum_{n=1}^{\infty} f(n)\ \leq\ \sum_{n=0}^{\infty} 2^{n}f(2^{n})\ \leq\ 2\sum_{n=1}^{\infty} f(n)$$

answered Feb 10 '18 at 09:47

user

162,563

Gimusi. Very nice! – Peter Szilas Feb 10 '18 at 10:41
@PeterSzilas It is a very useful test for series! – user Feb 10 '18 at 10:43

Guy Fsone · Answer 3 · 2018-02-10T10:07:40.540

1

Hint: By Cauchy condensation test your series is equivalent to

$$\sum_{n=2}^{\infty} \frac{2^n}{2^n(\ln (2^n))^p}=\frac{1}{\ln^p 2}\sum_{n=2}^{\infty} \frac{1}{n^p}$$ can you conclude?

edited Feb 10 '18 at 10:07

answered Feb 10 '18 at 09:46

Guy Fsone

25,237

Thanks for the answer, however we have to use the integral test since this exercise is part of an introductory real analysis course. – Dr. Heinz Doofenshmirtz Feb 10 '18 at 09:47
@HeinzDoofenschmirtz that is more easier – Guy Fsone Feb 10 '18 at 09:49
Guy. Very nice! – Peter Szilas Feb 10 '18 at 10:40
@PeterSzilas yeeaa very nice but with low votes so far – Guy Fsone Feb 10 '18 at 10:47
Guy. You please do not worry so much about votes.You are good at it, and I , and certainly many others, enjoy reading your nice posts.Keep it up! – Peter Szilas Feb 10 '18 at 10:50

Convergence of $\sum_{n=2}^\infty \frac{1}{(n(\log n)^p )}$ using the integral test

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