Test the series for convergence or divergence

Question

Test the series for convergence or divergence

(a) $$\sum_{n=2}^\infty{1\over{{(\ln n)}^{\ln n}}}$$

(b) $$\sum_{n=1}^\infty{(\sqrt[n]{2}-1)}$$

and I found out that these two are some what converging but don't know how to reason it. Please help!!

Now I think I got (b) so Please help with (a)!!!

Answer 1

Use this result,

Find the set of $x>0$ such that the series $\sum\limits_n x^{\ln{n}}$ converges

$\log n \ge 3 $, $n\ge 21 \implies$ $\frac{1}{\log n} \leq \frac{1}{3}\implies$ $ (\frac{1}{\log n})^{\log n} \leq {\frac{1}{3}}^{\log n}$. So, converges(Since, $ 3 >e$. So,$1/3<1/e)$