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This question already has an answer here:

Is the following series converging $\sum_{n=2}\dfrac{1}{(\ln n)^{\ln n}}$

I am not able to compare this with anything, can some show the way

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marked as duplicate by Brian Fitzpatrick, Jonas Meyer, saz, Hakim, Adam Hughes Oct 28 '14 at 19:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Hint: Our denominator is $\exp((\ln\ln n)(\ln n))$, which is $n^{\ln\ln n}$. Now think $p$-series and Comparison.

Remark: When we meet $a^b$, the fact that it is equal to $\exp((\ln a)(b))$ is frequently useful.

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  • $\begingroup$ Ok thinking like $p$- series we see that $ln ln n >1$ so can we conclude directly that the series diverges? Why we need comparison? $\endgroup$ – Kushal Bhuyan May 3 '16 at 2:23
  • $\begingroup$ Basically that's it, though early on $\ln\ln n\lt 1$. The term $p$-series is reserved for series $\sum\frac{1}{n^p}$ where $p$ is constant so ours is not a $p$-series, but after a while the $n$-th term is smaller than $\frac{1}{n^2}$ and we do comparison. $\endgroup$ – André Nicolas May 3 '16 at 3:42
  • $\begingroup$ Hmmm that's right. $\endgroup$ – Kushal Bhuyan May 3 '16 at 4:39
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HINT:
L'Hospitals rule gives us$$\lim_{n\to \infty}\dfrac{\ln n}{n}=0.$$ Therefore $$2\le\ln n\le n$$ for large $n$ values. $$\dfrac{1}{\ln n^{\ln n}}\le\dfrac{1}{2^n}.$$ I'm sure you can continue from here.

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  • $\begingroup$ I don't see how you got your last inequality from the preceding one. (Doesn't this imply that $n\ln 2\le (\ln n)(\ln(\ln n))$ for n large?) $\endgroup$ – user84413 Oct 29 '14 at 14:13
  • $\begingroup$ @user84413: $$\dfrac{1}{\ln n}\le\dfrac{1}{2}$$ $$(\dfrac{1}{\ln n})^{\ln n}\le(\dfrac{1}{2})^n$$ $\endgroup$ – Bumblebee Oct 30 '14 at 13:59
  • $\begingroup$ Thanks for your reply. I'm not sure this works, though. For example, $1/3\le1/2$, but $(\frac{1}{3})^2>(\frac{1}{2})^4$. $\endgroup$ – user84413 Oct 30 '14 at 15:06
  • $\begingroup$ @user84413: I see. Since $\frac{1}{2}<1,$ my second inequality is not true. I have to find a better way to do this. Thank you. $\endgroup$ – Bumblebee Oct 31 '14 at 3:26
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If $n\ge e^{(e^2)}$, then $\ln n\ge e^2\implies \ln(\ln n)\ge 2\implies$

$(\ln n)(\ln(\ln n))\ge 2 \ln n\implies \ln[(\ln n)^{\ln n}]\ge\ln(n^2)\implies(\ln n)^{\ln n}\ge n^2\implies \frac{1}{(\ln n)^{\ln n}}\le\frac{1}{n^2}$,

so the series converges by the Comparison Test with $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n^2}$.

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