# Test the series for convergence or divergence

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!!

• For (b), we have $\sqrt[n]2-1\sim\frac{\ln2}n$ – Kemono Chen May 31 '18 at 4:48
• First question. How did you write the first part with the sqrt part? – Applepie May 31 '18 at 4:50
• @Applepie, Are you sure the first one is convergent? – Topology May 31 '18 at 7:10

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,$1/3<1/e)$

• @Applepie please check the link and check whether my reasoning is right or not? – Unknown x May 31 '18 at 6:21
• umm isn't $\log n$ smaller than n? how can it be bigger? – Applepie May 31 '18 at 6:22
• Yes. sorry! Let me check any other bound. – Unknown x May 31 '18 at 6:25
• @Applepie How is my answer. right now? – Unknown x May 31 '18 at 6:29
• In your case doesn't n just need to be 3? since if you take n as above 3 you have $1\over n$ smaller than 1. If you have n to $\infty$ and if you have number below 1 it goes to 0. And I think it is right!( Well this is what I think... I don't have the answer.... sorry) – Applepie May 31 '18 at 6:37

Hint:

If $n > e^{e^2}$, then $$(\ln n)^{\ln n} = e ^{ \ln(\ln n) \ln n}= n ^{ \ln(\ln n)} > n^2$$