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Why is $$\sum \frac{1}{\ln(n)^{\ln n}}$$ convergent?

I tried Cauchy's condensation test along with other test but nothing works well on this series. Any ideas?

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    $\begingroup$ Hint: $(\log(n))^{\log(n)} = n^{\log(\log(n))}$ . $\endgroup$ – Dan Brumleve Jan 20 '18 at 18:09
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$\log(n)$ is unbounded, in particular for $N$ big enough $\log(n) > e^2$ for all $n\ge N$. Then

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

which converges, as you probably know.

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Rewrite the general term as $$\frac1{(\ln n)^{\ln n}}=\frac1{e^{\ln n\,\ln(\ln n)}}=\frac1{n^{\ln(\ln n)}}.$$ Now $\dfrac 1{n^{\scriptstyle \ln(\ln n)}}=o\Bigl(\dfrac1{n^r}\Bigr)$ for any $r>0$. In particular it is $\;o\Bigl(\dfrac1{n^2}\Bigr)$, which converges.

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