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This question already has an answer here:

How can I determine if $\sum_{n\geq1}\frac{1}{(\ln n)^{\ln n}}$, any ideas?

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marked as duplicate by Hans Lundmark, Misha Lavrov, Rebellos, Leucippus, user99914 Dec 10 '17 at 3:30

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  • $\begingroup$ Should your sum maybe start at $n=2$? Because if $n = 1$, then $\ln(1) = 0$ appears in the denominator. $\endgroup$ – TastyRomeo Nov 10 '16 at 10:48
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    $\begingroup$ $a^{\ln n} = n^{\ln a}$ $\endgroup$ – reuns Nov 10 '16 at 10:51
  • $\begingroup$ @SteamyRoot: Perhaps you can take $0^0=1$ as is done in combinatorics and so $\dfrac{1}{(\log_e 1)^{\log_e 1}}=1$ $\endgroup$ – Henry Nov 10 '16 at 10:59
  • $\begingroup$ About $5.717$ if you start the sum at $n=2$, so about $6.717$ if you start the sum at $n=1$ $\endgroup$ – Henry Nov 10 '16 at 11:06
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Notice we can write

$$ (\ln n)^{\ln n} = e^{ \ln n ( \ln ( \ln n ) ) } = n^{\ln ( \ln n) } $$

Notice $\ln ( \ln n) ) \to \infty $, so eventually $\ln (\ln n ) > 2 $, after say some $N$. Thus, after some $N$, we have

$$ n^{ \ln ( \ln n) } > n^2 $$

which implies that

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

Since user1952009 does not agree with me giving full answer, here is the full answer:

Now, since $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, then by the comparison test, we must have that

$$ \boxed{ \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{\ln n}} \; \; \; \; \bf{Converges } }$$

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  • $\begingroup$ you are allowed to let the OP search by himself (give some hints instead of the full answer) $\endgroup$ – reuns Nov 10 '16 at 10:53
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    $\begingroup$ Why the downvote? It would be nice if the downvoter explain why he downvoted so I can fix my answer $\endgroup$ – ILoveMath Nov 10 '16 at 10:54
  • $\begingroup$ I just did.${}{}$ $\endgroup$ – reuns Nov 10 '16 at 10:56

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