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How to determine convergence of the series: $$\sum_{n=1}^\infty\frac{1}{\ln(n)^{\ln(n)}}$$

I spent most of the time using the Integral criteria (since the function $f(x)=\frac{1}{\ln(x)^{\ln(x)}}$ is non increasing positive function) but didn't manage to get a solution. I used partial integration method, however it kept getting more and more complex.

Is there any better way of determining convergence of the series?

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  • $\begingroup$ Hint: For every $n$ large enough, $$(\ln n)^{\ln n}\geqslant n^2.$$ $\endgroup$
    – Did
    Jun 21 '15 at 22:02
  • $\begingroup$ By the way, $\frac{1}{\ln n^{\ln n}}$ is $\frac{1}{0^0}$. While I'm of the opinion that $0^0$ should be defined as $1$, enough people disagree. $\endgroup$ Jun 22 '15 at 0:04
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It converges. $$\log(x)^{\log x} = x^{\log\log x} $$ and for every $x$ big enough we have $\log\log x\geq 2$.

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