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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)!!!

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  • $\begingroup$ For (b), we have $\sqrt[n]2-1\sim\frac{\ln2}n$ $\endgroup$ – Kemono Chen May 31 '18 at 4:48
  • $\begingroup$ First question. How did you write the first part with the sqrt part? $\endgroup$ – Applepie May 31 '18 at 4:50
  • $\begingroup$ @Applepie, Are you sure the first one is convergent? $\endgroup$ – Topology May 31 '18 at 7:10
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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,$1/3<1/e)$

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  • $\begingroup$ @Applepie please check the link and check whether my reasoning is right or not? $\endgroup$ – Unknown x May 31 '18 at 6:21
  • $\begingroup$ umm isn't $\log n$ smaller than n? how can it be bigger? $\endgroup$ – Applepie May 31 '18 at 6:22
  • $\begingroup$ Yes. sorry! Let me check any other bound. $\endgroup$ – Unknown x May 31 '18 at 6:25
  • $\begingroup$ @Applepie How is my answer. right now? $\endgroup$ – Unknown x May 31 '18 at 6:29
  • $\begingroup$ 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) $\endgroup$ – Applepie May 31 '18 at 6:37
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Hint:

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

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