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As the title suggests, I'd like to prove that the sum $$ \sum_{n=2}^\infty (\ln n)^{- \ln n} $$ is finite. The root and ratio test both fail here, but WA suggests that there is a comparison that can be used to show convergence.

The only thought I have is that it may help to write the terms as $e^{-\ln(n)\ln(\ln(n))}$, but this has not led me to any particular insight. Any ideas are appreciated.

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Well,

$$e^{-(\log n)(\log \log n)} = n^{-\log \log n} < n^{-2}$$

for $n > e^{e^2}$. So a comparison with $\sum \frac{1}{n^2}$ shows the convergence.

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  • $\begingroup$ Can't believe I didn't see that. Thank you $\endgroup$ – Omnomnomnom Jun 20 '14 at 17:47
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There is a theorem that for a decreasing series $\sum a_n$ converges if and only if $\sum 2^k a_{2^k}$ converges. Applying that in this case gives, where I assume the log is base $2$ just to make my life easy, $$\sum \left(\frac{2}{k}\right)^k$$ and this is easily seen to converge by root or ratio.

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For large enough $n$, one has $\ln n > e^2$, so that $(\ln n)^{-\ln n} < e^{-2\ln n} = {1 \over n^2}$.

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By Cauchy condensation test we have

$$\sum 2^na_{2n}=\sum \frac{2^n}{(\log 2^n)^{\log 2^n}}=\sum \frac{2^n}{n^{(n\log 2)}\cdot{(\log2)}^{(n\log 2)}}$$

which converges for example by limit comparison test with $\sum \frac1{n^2}$.

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Hint: You can use Cauchy condensation test.

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