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How to test the convergence of following series $$\sum{\frac{1}{(\ln {n})^{\ln{n}}}}$$

I have tried Cauchy condensation test and gives me nothing


marked as duplicate by Clement C., Lucian, C. Falcon, user99914, Daniel W. Farlow Jan 23 '17 at 1:35

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    $\begingroup$ See this. $\endgroup$ – David Mitra Jan 22 '17 at 22:37
  • $\begingroup$ @DavidMitra Perfect find :D $\endgroup$ – Simply Beautiful Art Jan 22 '17 at 22:38
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    $\begingroup$ @SimplyBeautifulArt It was the second "related" question :) $\endgroup$ – David Mitra Jan 22 '17 at 22:39
  • $\begingroup$ @DavidMitra >.> Well at least I didn't use a totally copied answer from that place. $\endgroup$ – Simply Beautiful Art Jan 22 '17 at 22:40

Method 1: Claim: $(\ln n)^{\ln n} > n^2$ for $n>e^{e^2}.$ The claim implies

$$\frac{1}{(\ln n)^{\ln n}} < \frac{1}{n^2}$$

for these $n.$ Since $\sum 1/n^2$ converges, so does $\sum 1/(\ln n)^{\ln n}.$

To prove the claim, apply $\ln $ to both sides. I'll leave this to the reader for now.

Method 2: Cauchy condensation: Consider the series

$$\tag 1\sum \frac{2^n}{(\ln 2^n)^{\ln 2^n}} = \sum \frac{2^n}{(n\ln 2)^{n\ln 2} }= \sum \frac{2^n}{((n\ln 2)^{\ln 2})^n}.$$

Now $(n\ln 2)^{\ln 2}\to \infty,$ so eventually $(n\ln 2)^{\ln 2} \ge 4.$ This shows the terms in $(1)$ are eventually bounded above by $2^n/4^n = 1/2^n.$ This proves convergence of the original series.


Let's try that Cauchy condensation test again with a ratio test:

$$\frac1{(\ln n)^{\ln n}}\implies\frac{2^n}{(\ln2^n)^{\ln2^n}}=\frac{2^n}{(cn)^{cn}}\implies\frac{(2/c(n+1))^{c(n+1)}}{(2/cn)^{cn}}\to0$$

where $c=\ln2$.

Thus, it converges.

  • $\begingroup$ $\frac{2^n}{(\ln2^n)^{\ln2^n}}<\frac{2^n}{(\ln e^n)^{\ln e^n}}$ Are you sure it's ok, since $\ln{2^n}<\ln{e^n}$ ? $\endgroup$ – UfmdFkiF Jan 22 '17 at 22:42
  • $\begingroup$ @TheMeff Oh, whoops, that went wrong. But either way... $\endgroup$ – Simply Beautiful Art Jan 22 '17 at 22:43
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    $\begingroup$ Usually $\implies $ means "implies". $\endgroup$ – zhw. Jan 22 '17 at 23:10

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