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Is the series

$$\sum_{n=2}^\infty \frac{1}{(\log (n))^{\log (\log (n))}}$$

convergent?

I know that $ \log(\log n) >2$. Now $\log(\log(\log (n))) > \log 2$. After that I am not able to proceed further.

Mark Viola
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3 Answers3

5

Hint: $a^b$ $= (e^{\log a})^b$ $= e^{b\log a}$. Applying this to your denominator turns it into $e^{(\log\log n)^2}$, and now you should be able to compare with $e^{\log n}$...

Steven Stadnicki
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3

Note that since $x\ge \log^2(x)$ for $x\ge \log(2)$, we have

$$\begin{align} \int_2^L \frac{1}{(\log(x))^{\log(\log(x))}}\,dx&\overbrace{=}^{x\mapsto e^x}\int_{\log(2)}^{\log(L)}\frac{e^x}{x^{\log(x)}}\,dx\\\\ &=\int_{\log(2)}^{\log(L)}e^{x-\log^2(x)}\,dx\\\\ &\ge \log(L/2) \end{align}$$

Hence, the integral test guarantees that the series of interest diverges.

Mark Viola
  • 179,405
2

HINT

Let use the Cauchy condensation test

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

that is

$$\sum_{n=2}^\infty \frac{1}{(\log n)^{\log\log n}} \ge \frac12\sum_{n=2}^\infty \frac{2^n}{(\log 2^n)^{\log\log 2^n}}=\frac12\sum_{n=2}^\infty \frac{2^n}{(n\log 2)^{\log (n\log 2)}}$$

user
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  • im not getting @gimusi...is there is any simple method is there...? –  Apr 09 '18 at 19:12
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    @Stupid that's not so bad! – user Apr 09 '18 at 19:13
  • Why dont you finish the question ? – Rene Schipperus Apr 09 '18 at 19:17
  • @ReneSchipperus it's just an hint and the last series seems seriously aimed to diverge – user Apr 09 '18 at 19:20
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    @Stupid To finish note that $$\frac{2^n}{(n\log 2)^{\log (n\log 2)}}\ge\frac{2^n}{n^{\log (n)}}$$ – user Apr 09 '18 at 19:31
  • @user, $$;\dfrac{2^n}{n^{\log(n)}}=\dfrac{e^{n\log(2)}}{e^{\log^2(n)}}=e^{n\log(2)-\log^2(n)}>e^{\frac12n-16\log^2(\sqrt[4]n)}> e^{\frac12n-16\sqrt n}=e^{\frac12\sqrt n\left(\sqrt n-32\right)}>e^{\sqrt n};$$ for any $;n>34^2=1156,.,$ Is there a quicker way to get a divergent series whose terms are smaller than the original series ? – Angelo May 31 '23 at 21:13
  • @Angelo We can also use root test $\sqrt[n]{\dfrac{2^n}{n^{\log(n)}}} \to 2$ or from here $e^{n\log 2 -\log^2 n )}= e^{n\left(\log 2-\frac{\log^2 n}{n}\right)}$, in both cases using that $\frac{\log^2 n}{n} \to 0$. – user Jun 01 '23 at 08:33
  • @user, I know we can use root test or calculate a limit, but I ask you if there is a quicker way to get a divergent series whose terms are smaller than the original series (by using only inequalities). – Angelo Jun 01 '23 at 11:49
  • @Angelo isn’t this answer https://math.stackexchange.com/a/2729853 good enough? – user Jun 01 '23 at 12:14
  • @user, yes, of course it is a very good answer, but I want to know how to get, in a quick way, a divergent series smaller than the original one without using integral test or root test, etc, but using only inequalities. – Angelo Jun 01 '23 at 15:38
  • @Angelo Indeed in that solution they are suggesting $$\frac{1}{(\log (n))^{\log (\log (n))}}=\frac 1{e^{(\log\log n)^2}}\ge \frac 1{e^{\log n}}=\frac1n$$ Isn't it the kind of solution you are looking for? – user Jun 01 '23 at 17:18
  • @user, actually I am looking for a short solution in which we find a divergent series smaller than $\displaystyle\sum\limits_{n=1}^{\infty}\frac{2^n}{n^{\log(n)}}$ by using only inequalities. I found $\displaystyle\sum\limits_{n=1157}^{\infty}e^{\sqrt n}$ but my solution is not short and I do not like it. – Angelo Jun 01 '23 at 18:01
  • Your solution is fine, more simply it suffices observe that $$\frac{2^n}{n^{\log n}}=\frac{e^{n\log 2}}{e^{\log^2n}}\ge \frac{e^{\log^2 n}}{e^{\log^2n}}=1$$ – user Jun 01 '23 at 23:25
  • @Angelo Is this way good for what you are looking for? – user Jun 03 '23 at 11:10
  • @user, I am looking for something similar, but could you explain me how you have proved that $,n\log2\geqslant \log^2n,$ for any $,n\in\Bbb N;?$ – Angelo Jun 03 '23 at 11:53
  • It is essentially related to $e^n\ge n^a$ for any $a\in \mathbb R$. There are many proof of that result on the site. – user Jun 03 '23 at 12:29
  • @user, rather it is related to $e^n\geqslant n^{\log n}$. Please, prove that $n\log2\geqslant\log^2n$ for any $n\in\Bbb N$ by using only inequalities. – Angelo Jun 03 '23 at 12:32
  • @Angelo We have $n\ge \log^2n \iff \sqrt n\ge \log n \iff e^{\sqrt n}\ge n$. It is essentially the standard inequality I’ve indicated. – user Jun 03 '23 at 13:10
  • @Angelo Here there are several proof https://math.stackexchange.com/q/55468 – user Jun 03 '23 at 17:10
  • It seems that there are not proofs shorter than the one I wrote in a previous comment. Moreover $,n\log 2\geqslant\log^2n,$ is not equivalent to $,n\geqslant\log^2n,$ or to $,e^{\sqrt n}\geqslant n$. – Angelo Jun 03 '23 at 18:06
  • @Angelo As I said from the begining you proof is fine, then you have asked for an alternative and I've given all the example I can. You can also make an official answer here or in the linked question. Regarding the inequality note that $$e^{\sqrt n}\ge n \implies \sqrt n \ge \log n \implies n\ge \log^2 n$$ the factor $\log 2$ is not relevant since in the same way we can prove that $e^{\sqrt n}\ge n^2$. Here is another related question where you could add your answer https://math.stackexchange.com/q/1287704/505767 – user Jun 03 '23 at 18:26
  • @user, thank you very much. I really appreciate all the help you have given me. – Angelo Jun 03 '23 at 19:13
  • @user, I have already added my answer. Click here. – Angelo Jun 03 '23 at 20:30