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Let $u_n=\frac{1}{(\log n)^{\log n}}$ then what can we say about convergence of the sequence and series as well... I tried Cauchy condensation formula but that doesn't seems to work...

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Hint:

$$(\log n)^{\log n} = n^{\log \log n}$$

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Well, let me give the condensation formula a try...

$$u_n=\frac{1}{(\log n)^{\log n}}$$

$$v_n=\frac{2^n}{(n\log 2)^{n\log 2}} = \left(\frac{2}{(n\log 2)^{\log 2}}\right)^n$$

Now use the root test:

$$\lim_{n\to\infty}\sqrt[n]{\left(\frac{2}{(n\log 2)^{\log 2}}\right)^n} = \lim_{n\to\infty}\frac{2}{(n\log 2)^{\log 2}} = \frac{2}{(\log 2)^{\log 2}}\lim_{n\to\infty}n^{-\log 2} = 0 < 1$$

So it converges.

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