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...
Convergence of the sequence
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$\begingroup$
convergence
2 Answers
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Hint:
$$(\log n)^{\log n} = n^{\log \log n}$$
0
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.