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I'm having some trouble approaching this problem. "For what values of $p\in\mathbb{R}$ does the series $$\sum^{\infty}_{n=4}{\frac{1}{n\log (n)\log( \log(n))^p}}$$ converge?" For a fixed p, I could see approaching this with some of the standard tests for convergence but I am unsure how to find p. Any answers or hints would be appreciated, thanks!

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Do you know the corresponding answer (more importantly, technique) for $\sum \frac{1}{n (\log n)^p}$? – Qiaochu Yuan Aug 7 '11 at 1:02
Hint: Integral test – André Nicolas Aug 7 '11 at 1:47
Hint: What’s the derivative with respect to $x$ of $\log(\log(x))$? – Brian M. Scott Aug 7 '11 at 1:48
If you have the Cauchy condensation test, you can apply it twice and you'll have the answer. – Zarrax Aug 7 '11 at 3:01

You can prove convergence using the integral test. This holds only for monotonically decreasing functions, so you should first convince yourself that $f(x)=\frac{1}{x\log (x)\log( \log x)^p}$ is monotonically decreasing on $[4,\infty)$. First, note that since $x \geq 4$, the functions $x$, $\log(x)$ and $\log(\log x)$ are positive and monotonically increasing. Then for $p \geq 0$, $x\log (x)\log( \log x)^p$ is clearly increasing, so $f(x)$ is monotonically decreasing. For $p < 0$, then $\frac{1}{x\log (x)\log( \log x)^p}=\frac{\log( \log x)^{|p|}}{x\log (x)}$. If $|p|$ were large enough for the numerator to grow faster than the denominator, then our sum diverges trivially. Otherwise, the denominator grows faster and so $f(x)$ is again monotonically decreasing.

Now, we apply the integral test. Our sum converges iff $\int_4^\infty f(x)dx$ converges. Using a simple u-substitution ($u=\log(\log x))$), we get $$\int_4^\infty f(x)dx = \int_{\log(\log4)}^\infty u^{-p} du = \frac{u^{1-p}}{1-p} |_{\log(\log4)}^\infty$$ which only converges for $p > 1$.

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The integral test works here. And to check your answer, it works for the same p that $\displaystyle \sum \frac{1}{n (\log n) ^ p}$ converges.

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Oh, I really don't think that any comments were there when I posted this answer. I missed them entirely - but it's still true. – mixedmath Aug 7 '11 at 2:00
Maybe this answer should be CW :) – The Chaz 2.0 Aug 7 '11 at 2:49
@The Chaz Consider it done! – mixedmath Aug 7 '11 at 3:47
My most popular answers are the ones I just mashed together from comments! – The Chaz 2.0 Aug 7 '11 at 4:22

I realise this question is ancient, but IMO, integral test is not the most straightforward way of doing this type of series. The best test is Cauchy condensation test. I have written up an answer here:

Convergence of series involving in iterated logarithms $\sum \frac{1}{n(\log n)^{\alpha_1}\cdots (\log^k(n))^{\alpha_k} }$

This test reduce higher iterations of log and it is a good tool in the bag. It is basically the intuition behind the elementary proof of divergence of harmonic series.

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Interesting!!!! – Tony Jun 25 '14 at 23:48
@user151558 thanks though this is a very old post.... note the question was asked 3 years ago :P – Lost1 Jun 25 '14 at 23:59
I know, I just wanted to make sure you knew that even these old ones don't go unnoticed forever :).(You had no comments or upvotes for 4 months) – Tony Jun 26 '14 at 0:51

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