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

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 $\sum_{n=1}^\infty a_n<\infty \iff \int_1^\infty f(x)\,dx<\infty$. – Heinz Doofenschmirtz Feb 10 '18 at 9:45
• @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. – Masacroso Feb 10 '18 at 9:47
• – Guy Fsone Feb 10 '18 at 10:04

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 $p$.

• 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! – Heinz Doofenschmirtz Feb 10 '18 at 9:51

You can use the Cauchy condensation test

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

• 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

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?

• Thanks for the answer, however we have to use the integral test since this exercise is part of an introductory real analysis course. – Heinz Doofenschmirtz Feb 10 '18 at 9:47
• @HeinzDoofenschmirtz that is more easier – Guy Fsone Feb 10 '18 at 9:49
• 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